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SYLLABUS
Measurement of angles – Degree and Radian – Relation between degree and radian – Definition of trigonometric ratios – Trigonometric ratios of standard angles – Graphs of 𝑠𝑖𝑛 𝑥 , 𝑐𝑜𝑠 𝑥 ,𝑡𝑎𝑛 𝑥 and 𝑒𝑥 – Compound angles – Expansions of 𝑠𝑖𝑛 (𝐴±𝐵) , 𝑐𝑜𝑠 (𝐴±𝐵) , 𝑡𝑎𝑛 (𝐴±𝐵) (without proof) – Trigonometric ratios of multiple angles of 2𝐴 – Simple problems – Engineering applications of Trigonometry.
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\[\color {royalblue} {Introduction}:\ \hspace{20cm}\]
\[Trigonometry\ is\ one\ of\ the\ oldest\ branches\ of\ Mathematics.\ The\ word\]
\[Trigonometry\ is\ derived\ from\ the\ Greek\ words\ ‘Trigonon’\ and\ ‘metron’\ means\]
\[measurement\ of\ angles.\ In\ olden\ days\ Trigonometry\ was\ mainly\ used\ as\ a\ tool\ for\]
\[studying\ astronomy.\ In\ earlier\ stages\ Trigonometry\ was\ mainly\ concerned\ with\ angles\]
\[of\ a\ triangle.\ But\ now\ it\ has \ its\ applications\ in\ various\ branches\ of\ science\ such\ as\]
\[surveying,\ engineering,\ navigations\ etc.\ For\ the\ study\ of\ higher\ mathematics,\ knowledge\ of\ Trigonometry\ is\ essential.\]
\[\color {royalblue} {Measurement\ of\ angles}:\ \hspace{20cm}\]
Two types of units of measurements of an angle which mostly commonly used namely Degree measure and Radian measure.
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Radian measure is more convenient for analysis whereas degree measure of an angle is more convenient to communicate the concept between people.
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\[\color {brown} {Degree\ Measure\ :}\ \hspace{20cm}\]
\[If\ a\ rotation\ from\ the\ initial\ side\ to\ terminal\ side\ is\ (\frac{1}{360})^{th}\ of\ a\ revolution,\]\[ the\ angle\ is\ said\ to\ have\ a\ measure\ of\ one\ degree,\ written\ as\ 1^o\]
\[\color {brown} {Radian\ Measure\ :}\ \hspace{20cm}\]
The radian measure of an angle is the ratio of the arc length it subtends to the radius of the circle in which it is the central angle.
\[covert\ radians\ into\ degrees\ ———–\ radians\ \times\ \frac{180^0}{\pi}\]
\[covert\ degrees\ into\ radians\ ———–\ degrees\ \times\ \frac{\pi\ radians}{180^0}\]
\[\color {purple} {Example\ 1:}\ \color {red} {Convert}\ 240^0\ to\ equivalent\ radians\ value\ \hspace{15cm}\]
\[\hspace{5cm}\ October\ 2023\]
\[\color {blue}{Solution:}\ \hspace{20cm}\]
\[\hspace{2cm}\ 240^0\ \times \frac{\pi\ radians}{180^0}\]
\[=\ \hspace{2cm}\ \frac{4}{3}\ \pi\ radians\]
\[\color {purple} {Example\ 2:}\ \color {red} {Convert}\ 30^0\ to\ radians\ \hspace{15cm}\]
\[\hspace{5cm}\ April\ 2024\]
\[\color {blue}{Solution:}\ \hspace{20cm}\]
\[\hspace{2cm}\ 30^0\ \times \frac{\pi\ radians}{180^0}\]
\[=\ \hspace{2cm}\ \frac{\pi}{6}\ radians\]
\[\color {purple} {Example\ 3:}\ \color {red} {Convert}\ 60^0\ to\ radians\ \hspace{15cm}\]
\[\hspace{5cm}\ April\ 2025\]
\[\color {blue}{Solution:}\ \hspace{20cm}\]
\[\hspace{2cm}\ 60^0\ \times \frac{\pi\ radians}{180^0}\]
\[=\ \hspace{2cm}\ \frac{\pi}{3}\ radians\]
\[\color {purple} {Example\ 4:}\ \color {red} {Convert}\ \frac{\pi}{4}\ in\ to\ degrees\ \hspace{15cm}\]
\[\hspace{5cm}\ October\ 2024\ June\ (Supp)\ 2025\]
\[\color {blue}{Solution:}\ \hspace{20cm}\]
\[\hspace{2cm}\ \frac{\pi}{4}\ \times \frac{180^0}{\pi}\]
\[=\ \hspace{2cm}\ 45^0\]
\[\color {royalblue} {Trigonometrical\ ratios}:\ \hspace{20cm}\]
\[There\ are\ six\ Trigonometrical\ ratios\ sine,\ cosine,\ tangent,\ cotangent,\ secant\ and\]\[cosecant\ shortly\ written\ as\ sin θ,\ cos θ,\ tan θ,\ cot θ,\ sec θ\ and\ cosec θ.\]
\[\color {royalblue} {Fundamental\ Trigonometrical\ identities}:\ \hspace{20cm}\]
\[1\ .\ sin^2θ\ +\ cos^2θ\ =\ 1\ \hspace{10cm}\]
\[2\ .\ 1\ +\ tan^2θ\ =\ sec^2θ\ \hspace{10cm}\]
\[3\ .\ 1\ +\ cot^2θ\ =\ cosec^2θ\ \hspace{10cm}\]
\[\color {purple} {Example\ 5:}\ If\ cos\ \theta\ =\ \frac{3}{5},\ then\ \color {red} {find}\ \text{the values of other five trigonometric}\ \hspace{10cm}\]\[ratios\ \hspace{15cm}\]
\[\hspace{5cm}\ October\ 2023\ April\ 2025\]
\[\color {blue}{Solution:}\ \hspace{20cm}\]
\[\hspace{2cm}\ Given\ cos\ \theta\ =\ \frac{3}{5}\ \hspace{18cm}\]
\[\hspace{2cm}\ sin^2θ\ +\ cos^2θ\ =\ 1\ \hspace{10cm}\]
\[\hspace{2cm}\ sin^2θ\ =\ 1\ -\ cos^2θ\ \hspace{10cm}\]
\[\hspace{3cm}\ =\ 1\ -\ (\frac{3}{5})^2\ \hspace{10cm}\]
\[\hspace{3cm}\ =\ 1\ -\ \frac{9}{25}\ \hspace{10cm}\]
\[\hspace{3cm}\ =\ \frac{25\ -\ 9}{25}\ \hspace{10cm}\]
\[\hspace{3cm}\ =\ \frac{16}{25}\ \hspace{10cm}\]
\[\hspace{2cm}\ \boxed{sin\ \theta\ =\ \pm\ \frac{4}{5}}\ \hspace{10cm}\]
\[\hspace{2cm}\ tan\ θ\ =\ \frac{sin\ \theta}{cos\ \theta}\ \hspace{10cm}\]
\[\hspace{3cm}\ =\ \frac{\pm\ \frac{4}{5}}{\frac{3}{5}}\ \hspace{10cm}\]
\[\hspace{2cm}\ \boxed{tan\ \theta\ =\ \pm\ \frac{4}{3}}\ \hspace{10cm}\]
\[\hspace{2cm}\ cot\ θ\ =\ \frac{1}{tan\ \theta}\ \hspace{10cm}\]
\[\hspace{3cm}\ =\ \frac{1}{\pm\ \frac{4}{3}}\ \hspace{10cm}\]
\[\hspace{2cm}\ \boxed{cot\ \theta\ =\ \pm\ \frac{3}{4}}\ \hspace{10cm}\]
\[\hspace{2cm}\ sec\ θ\ =\ \frac{1}{cos\ \theta}\ \hspace{10cm}\]
\[\hspace{3cm}\ =\ \frac{1}{\frac{3}{5}}\ \hspace{10cm}\]
\[\hspace{2cm}\ \boxed{sec\ \theta\ =\ \frac{5}{3}}\ \hspace{10cm}\]
\[\hspace{2cm}\ cosec\ θ\ =\ \frac{1}{sin\ \theta}\ \hspace{10cm}\]
\[\hspace{3cm}\ =\ \frac{1}{\pm\ \frac{4}{5}}\ \hspace{10cm}\]
\[\hspace{2cm}\ \boxed{sec\ \theta\ =\ \pm\ \frac{5}{4}}\ \hspace{10cm}\]
\[\color {purple} {Example\ 6:}\ If\ cos\ \theta\ =\ \frac{5}{13},\ then\ \color {red} {find}\ \text{the values of other five trigonometric}\ \hspace{10cm}\]\[ratios\ \hspace{15cm}\]
\[\hspace{5cm}\ April\ 2024\]
\[\color {blue}{Solution:}\ \hspace{20cm}\]
\[\hspace{2cm}\ Given\ cos\ \theta\ =\ \frac{5}{13}\ \hspace{18cm}\]
\[\hspace{2cm}\ sin^2θ\ +\ cos^2θ\ =\ 1\ \hspace{10cm}\]
\[\hspace{2cm}\ sin^2θ\ =\ 1\ -\ cos^2θ\ \hspace{10cm}\]
\[\hspace{3cm}\ =\ 1\ -\ (\frac{5}{13})^2\ \hspace{10cm}\]
\[\hspace{3cm}\ =\ 1\ -\ \frac{25}{169}\ \hspace{10cm}\]
\[\hspace{3cm}\ =\ \frac{169\ -\ 25}{169}\ \hspace{10cm}\]
\[\hspace{3cm}\ =\ \frac{144}{13}\ \hspace{10cm}\]
\[\hspace{2cm}\ \boxed{sin\ \theta\ =\ \pm\ \frac{12}{13}}\ \hspace{10cm}\]
\[\hspace{2cm}\ tan\ θ\ =\ \frac{sin\ \theta}{cos\ \theta}\ \hspace{10cm}\]
\[\hspace{3cm}\ =\ \frac{\pm\ \frac{12}{13}}{\frac{5}{13}}\ \hspace{10cm}\]
\[\hspace{2cm}\ \boxed{tan\ \theta\ =\ \pm\ \frac{12}{5}}\ \hspace{10cm}\]
\[\hspace{2cm}\ cot\ θ\ =\ \frac{1}{tan\ \theta}\ \hspace{10cm}\]
\[\hspace{3cm}\ =\ \frac{1}{\pm\ \frac{12}{5}}\ \hspace{10cm}\]
\[\hspace{2cm}\ \boxed{cot\ \theta\ =\ \pm\ \frac{5}{12}}\ \hspace{10cm}\]
\[\hspace{2cm}\ sec\ θ\ =\ \frac{1}{cos\ \theta}\ \hspace{10cm}\]
\[\hspace{3cm}\ =\ \frac{1}{\frac{5}{13}}\ \hspace{10cm}\]
\[\hspace{2cm}\ \boxed{sec\ \theta\ =\ \frac{13}{5}}\ \hspace{10cm}\]
\[\hspace{2cm}\ cosec\ θ\ =\ \frac{1}{sin\ \theta}\ \hspace{10cm}\]
\[\hspace{3cm}\ =\ \frac{1}{\pm\ \frac{12}{13}}\ \hspace{10cm}\]
\[\hspace{2cm}\ \boxed{sec\ \theta\ =\ \pm\ \frac{13}{12}}\ \hspace{10cm}\]
\[\color {purple} {Example\ 7:}\ If\ sin\ \theta\ =\ \frac{5}{13},\ \text{then find the values of other five trigonometric}\ \hspace{10cm}\]\[ratios\ \hspace{15cm}\]
\[\hspace{5cm}\ October\ 2024,\ June(Supp)\ 2025\]
\[\color {blue}{Solution:}\ \hspace{20cm}\]
\[\hspace{2cm}\ Given\ sin\ \theta\ =\ \frac{5}{13}\ \hspace{18cm}\]
\[\hspace{2cm}\ sin^2θ\ +\ cos^2θ\ =\ 1\ \hspace{10cm}\]
\[\hspace{2cm}\ cos^2θ\ =\ 1\ -\ cos^2θ\ \hspace{10cm}\]
\[\hspace{3cm}\ =\ 1\ -\ (\frac{5}{13})^2\ \hspace{10cm}\]
\[\hspace{3cm}\ =\ 1\ -\ \frac{25}{169}\ \hspace{10cm}\]
\[\hspace{3cm}\ =\ \frac{169\ -\ 25}{169}\ \hspace{10cm}\]
\[\hspace{3cm}\ =\ \frac{144}{13}\ \hspace{10cm}\]
\[\hspace{2cm}\ \boxed{cos\ \theta\ =\ \pm\ \frac{12}{13}}\ \hspace{10cm}\]
\[\hspace{2cm}\ tan\ θ\ =\ \frac{sin\ \theta}{cos\ \theta}\ \hspace{10cm}\]
\[\hspace{3cm}\ =\ \frac{\pm\ \frac{5}{13}}{\frac{12}{13}}\ \hspace{10cm}\]
\[\hspace{2cm}\ \boxed{tan\ \theta\ =\ \pm\ \frac{5}{12}}\ \hspace{10cm}\]
\[\hspace{2cm}\ cot\ θ\ =\ \frac{1}{tan\ \theta}\ \hspace{10cm}\]
\[\hspace{3cm}\ =\ \frac{1}{\pm\ \frac{5}{12}}\ \hspace{10cm}\]
\[\hspace{2cm}\ \boxed{cot\ \theta\ =\ \pm\ \frac{12}{5}}\ \hspace{10cm}\]
\[\hspace{2cm}\ sec\ θ\ =\ \frac{1}{cos\ \theta}\ \hspace{10cm}\]
\[\hspace{3cm}\ =\ \pm\ \frac{1}{\frac{12}{13}}\ \hspace{10cm}\]
\[\hspace{2cm}\ \boxed{sec\ \theta\ =\ \pm\ \frac{13}{12}}\ \hspace{10cm}\]
\[\hspace{2cm}\ cosec\ θ\ =\ \frac{1}{sin\ \theta}\ \hspace{10cm}\]
\[\hspace{3cm}\ =\ \frac{1}{\frac{5}{13}}\ \hspace{10cm}\]
\[\hspace{2cm}\ \boxed{sec\ \theta\ =\ \frac{13}{5}}\ \hspace{10cm}\]
\[\color {royalblue} {Trigonometrical\ ratios\ of\ known\ angles}:\ \hspace{20cm}\]
\[\color {royalblue} {Signs\ of\ Trigonometrical\ ratios}:\ \hspace{20cm}\]
| Quadrant | Sign of ratios | Remember |
| I | All are +ve ( 90 – θ, 360 + θ ) | All |
| II | Sin and cosec are +ve, Other ratios are –ve ( 90 + θ, 180 – θ ) | Silver |
| III | tan θ and cot θ are +ve, Other ratios are –ve ( 180 + θ, 270 – θ ) | Tea |
| IV | cos θ and sec θ are +ve, Other ratios are –ve ( 180 + θ, 360 – θ ) | Cups |
\[\color {brown} {Working\ rule\ for\ 180 ± θ\ and\ 360 ± θ}:\ \hspace{20cm}\]
| Ratio | Falls in Quadrant | Change of ratio |
| Sin (180 + θ ) | III | -Sin θ |
| Sin (180 – θ ) | II | Sin θ |
| Cos (180 + θ ) | III | – Cos θ |
| Cos (180 – θ ) | II | – Cos θ |
| Tan (180 + θ ) | III | Tan θ |
| Tan (180 – θ ) | II | – Tan θ |
| Sin (360 – θ ) | IV | – Sin θ |
\[\color {brown} {Working\ rule\ for\ 90 ± θ}:\ \hspace{20cm}\]
| Ratio | Falls in Quadrant | Change of ratio |
| Sin (90 + θ ) | II | Cos θ |
| Sin (90 – θ ) | I | Cos θ |
| cos (90 + θ ) | II | -Sin θ |
| cos (90 – θ ) | I | Cos θ |
\[\color {purple} {Example\ 8:}\ \color {red} {Show\ that}\ tan (765^0)\ =\ 1\ \hspace{15cm}\]
\[\color {blue}{Solution:}\ \hspace{20cm}\]
\[tan(765^0)\ =\ tan(360^0\ +\ 405^0)\ \hspace{18cm}\]
\[=\ tan(405^0)\ \hspace{18cm}\]
\[=\ tan(360^0\ +\ 45^0)\ \hspace{18cm}\]
\[=\ tan(45^0)\ \hspace{18cm}\]
\[=\ 1\ \hspace{18cm}\]
\[\boxed{Hence,\ tan (765^0)\ =\ 1}\]
\[\color {royalblue} {Graphs\ of\ Trigonometric\ functions}:\ \hspace{18cm}\]
\[\color {royalblue} {The\ Graph\ of\ Sine\ function}:\ \hspace{18cm}\]
\[\hspace{1cm}\ \text{To obtain the graph of the sine function, we begin by making a table of values of x and y that}\]
\[\text{satisfy the equation y = sin x}.\ \hspace{15cm}\]
\[\hspace{1cm}\ \text{Graphing each ordered pair (x, y) and then connecting them with a smooth curve, we obtain the}\]
\[\text{graph of y = sin x}\ \hspace{18cm}\]
\[\hspace{1cm}\ \text{The above graph represents only part of the graph of y = sin x. Since the sine function}\]
\[\text{is periodic, the graph continues in the same pattern in both directions. This graph is called as sine}\]
\[\text{wave or sinusoidal wave}.\ \hspace{18cm}\]
\[\color {purple} {characteristics\ of\ the\ function\ y\ =\ sin\ x}\ \hspace{15cm}\]
\[\color {purple} {Example\ 9:}\ \color {red} {Write}\ any\ two\ characteristics\ of\ the\ function\ y\ =\ sin\ x\ \hspace{15cm}\]
\[\hspace{5cm}\ April\ 2024\]
\[\color {blue}{Solution:}\ \hspace{20cm}\]
\[\hspace{2cm}\ 1.\ \textbf{Periodicity}\ \hspace{15cm}\]
\[\hspace{2cm}\ The\ sine\ function\ is\ periodic\ with\ a\ period\ of\ 2\ \pi.\ This\ means\ that\ the\ \hspace{2cm}\\ function\ repeats\ its\ values\ every\ 2\ \pi\ nuts.\ Mathematically\ sin(x\ +\ 2\ \pi)\ =\ sin(x)\ for\ all\ values\ of\ x.\]
\[\hspace{2cm}\ 2.\ \textbf{Range and amplitude}\ \hspace{15cm}\]
\[\hspace{2cm}\ \text{The range of the sine function is the set of all possible values it can}\ \hspace{2cm}\\ \hspace{2cm}\ \text{take, which is [1, -1]. This means that sinx oscillates between -1 and}\ \hspace{2cm}\\\hspace{2cm}\ \text{1. The amplitude, which is the maximum absolute value of the function, is 1.}\]
\[\color {royalblue} {The\ Graph\ of\ Cosine\ function}:\ \hspace{18cm}\]
\[\hspace{1cm}\ \text{To obtain the graph of the sine function, we begin by making a table of values of x and y that}\]
\[\text{satisfy the equation y = cos x}.\ \hspace{15cm}\]
\[\hspace{1cm}\ \text{Graphing each ordered pair (x, y) and then connecting them with a smooth curve, we obtain the}\]
\[\text{graph of y = cos x}\ \hspace{18cm}\]
\[\hspace{1cm}\ \text{The above graph represents only part of the graph of y = cos x. Since the cosine function}\]
\[\text{is periodic, the graph continues in the same pattern in both directions. This graph is called as sine}\]
\[\text{wave or cosine wave}.\ \hspace{18cm}\]
\[\color {purple} {characteristics\ of\ the\ function\ y\ =\ cos\ x}\ \hspace{15cm}\]
\[\color {purple} {Example\ 10:}\ \color {red} {Write}\ any\ two\ characteristics\ of\ the\ function\ y\ =\ cos\ x\ \hspace{15cm}\]
\[\hspace{5cm}\ April\ 2025\]
\[\color {blue}{Solution:}\ \hspace{20cm}\]
\[\hspace{2cm}\ 1.\ \textbf{Periodicity}\ \hspace{15cm}\]
\[\hspace{2cm}\ The\ sine\ function\ is\ periodic\ with\ a\ period\ of\ 2\ \pi.\ That\ means\ \hspace{2cm}\\ Mathematically\ cos(x\ +\ 2\ \pi)\ =\ cos(x)\ for\ all\ values\ of\ x.\]
\[\hspace{2cm}\ 2.\ \textbf{Range}\ \hspace{15cm}\]
\[\hspace{2cm}\ \text{The range of the cos x is }\ -1 \leq \cos x \leq 1
\hspace{2cm}\\ \hspace{2cm}\ \text{This means that cos x oscillates between -1 and 1}\ \hspace{2cm}\\\hspace{2cm}\]
\[\color {royalblue} {The\ Graph\ of\ the\ tanget\ function}:\ \hspace{18cm}\]
\[\hspace{1cm}\ \text{To obtain the graph of the tangent function, we begin by making a table of values of x and y that}\]
\[\text{satisfy the equation y = tan x}.\ \hspace{15cm}\]
\[\hspace{1cm}\ \text{Graphing each ordered pair (x, y) and then connecting them with a smooth curve, we obtain the}\]
\[\text{graph of y = tan x. As x approaches}\ \frac{\pi}{2}\ \text{from the left, the values of tan x approaches}\]
\[\text{positive infinity, As x approaches}\ \frac{\pi}{2}\ \text{from the right, the values of tan x approaches negative infinity}.\]
\[\hspace{1cm}\ \text{The above graph represents only part of the graph of y = tan x. The tangent function}\]
\[\text{is periodic with period π; the graph continues in the same pattern in both directions}.\ \hspace{2cm}\]
\[\color {purple} {characteristics\ of\ the\ function\ y\ =\ tan\ x}\ \hspace{15cm}\]
\[\color {royalblue} {The\ Graph\ of\ the\ exponential\ function}:\ \hspace{18cm}\]
\[\hspace{1cm}\ \text{In the exponential function}\ y\ =\ e^x\, \text{the base e is the natural exponential, being the number}\]
\[\text{ap- proximately equal to 2.718 and it is called as Euler’s number.}.\ \hspace{15cm}\]
\[\hspace{1cm}\ \text{Graphing each ordered pair (x, y) and then connecting them with a smooth curve, we obtain the}\]
\[\text{graph of y\ =\ e^x}\ \hspace{18cm}\]
\[\color {purple} {characteristics\ of\ the\ function\ y\ =\ e^x}\ \hspace{15cm}\]
\[\color {purple} {Example\ 11:}\ \color {red} {Write}\ any\ two\ characteristics\ of\ the\ function\ y\ =\ e^x\ \hspace{15cm}\]
\[\hspace{5cm}\ October\ 2023,\ October\ 2024\ June(Supp)\ 2025\]
\[\color {blue}{Solution:}\ \hspace{20cm}\]
\[\hspace{2cm}\ 1.\ \textbf{Exponential Growth}\ \hspace{15cm}\]
\[\hspace{3cm}\ The\ function\ y\ =\ e^x\ exhibits\ exponential\ growth.\ As\ x\ increases,\ y\ increases\ very\ rapidly\]
\[\hspace{3cm}\ Conversely,\ as x\ decreases,\ y\ approaches\ zero\ but\ never\ reaches\ it\]
\[\hspace{2cm}\ 2.\ \textbf{Derivative and Integral properties}\ \hspace{15cm}\]
\[\hspace{3cm}\ The\ function\ y\ =\ e^x\ \text{is unique in that it is its own derivative and integral}\]
\[\hspace{3cm}\ \frac{d}{dx}\ e^x\ =\ e^x\ and\ \int{e^x}\ dx\ =\ e^x\ +\ c\]
\[\color {purple} {Compound\ angles}:\ \hspace{15cm}\]
\[\color {royalblue} {Formulae}:\ \hspace{20cm}\]
\[1)\ Sin ( A + B )\ =\ Sin A\ Cos B\ +\ Cos A\ Sin B\]
\[2)\ Sin ( A – B )\ =\ Sin A\ Cos B\ -\ Cos A\ Sin B\]
\[3)\ Cos( A + B )\ =\ Cos A\ Cos B\ -\ Sin A Sin B\]
\[4)\ Cos( A – B )\ =\ Cos A\ Cos B\ +\ Sin A Sin B\]
\[5)\ Tan(A + B)\ =\ \frac{Tan A\ +\ Tan B}{1\ -\ Tan A\ Tan B}\]
\[6)\ Tan(A – B)\ =\ \frac{Tan A\ -\ Tan B}{1\ +\ Tan A\ Tan B}\]
\[\color {brown} {Note}:\ Sin\ ( – θ )\ =\ -\ sin\ θ\ \hspace{2cm}\ cos\ ( – θ )\ =\ cos\ θ\]
\[\color {purple} {Example\ 12:}\ \color {red} {Find\ the\ value\ of}\ sin\ {40}^0\ cos\ {20}^0\ +\ cos\ {40}^0\ sin\ {20}^0\ \hspace{15cm}\]
\[\hspace{5cm}\ April\ 2024,\ April\ 2025\]
\[\color {blue}{Solution:}\ W.\ K.\ T\ Sin ( A + B )\ =\ Sin A\ Cos B\ +\ Cos A\ Sin B\ \hspace{18cm}\]
\[sin\ {40}^0\ cos\ {20}^0\ +\ cos\ {40}^0\ sin\ {20}^0\ =\ Sin\ ({40}^0\ + {20}^0)\ \hspace{10cm}\]
\[=\ Sin \ {60}^0\ \hspace{10cm}\]
\[=\ \frac{\sqrt{3}}{2}\ \hspace{10cm}\]
\[\color{green}{\boxed{\therefore\ sin\ {40}^0\ cos\ {20}^0\ +\ cos\ {40}^0\ sin\ {20}^0\ =\ \frac{\sqrt{3}}{2}}}\]
\[\color {purple} {Example\ 13:}\ \color {red} {Find\ the\ value\ of}\ sin\ {50}^0\ cos\ {40}^0\ +\ cos\ {50}^0\ sin\ {40}^0\ \hspace{15cm}\]
\[\color {blue}{Solution:}\ W.\ K.\ T\ Sin ( A + B )\ =\ Sin A\ Cos B\ +\ Cos A\ Sin B\ \hspace{18cm}\]
\[sin\ {50}^0\ cos\ {40}^0\ +\ cos\ {50}^0\ sin\ {40}^0\ =\ Sin\ ({50}^0\ + {40}^0)\ \hspace{10cm}\]
\[=\ Sin \ {90}^0\ \hspace{10cm}\]
\[=\ 1\ \hspace{10cm}\]
\[\boxed{\therefore\ sin\ {50}^0\ cos\ {40}^0\ +\ cos\ {50}^0\ sin\ {40}^0\ =\ 1}\]
\[\color {purple} {Example\ 14:}\ \color {red} {Find\ the\ value\ of}\ cos\ {40}^0\ cos\ {20}^0\ -\ sin\ {40}^0\ sin\ {20}^0\ \hspace{15cm}\]
\[\color {blue}{Solution:}\ W.\ K.\ T\ Cos A\ Cos B\ -\ Sin A Sin B\ =\ Cos( A + B )\ \hspace{18cm}\]
\[cos\ {40}^0\ cos\ {20}^0\ -\ sin\ {40}^0\ sin\ {20}^0\ =\ cos\ ({40}^0\ + {20}^0)\ \hspace{10cm}\]
\[=\ Cos\ {60}^0\ \hspace{10cm}\]
\[=\ \frac{1}{2}\ \hspace{10cm}\]
\[\boxed{\therefore\ cos\ {40}^0\ cos\ {20}^0\ -\ sin\ {40}^0\ sin\ {20}^0\ =\ \frac{1}{2}}\]
\[\color {purple} {Example\ 15:}\ \color {red} {Evaluate}\ cos\ {75}^0\ \hspace{18cm}\]
\[\hspace{5cm}\ October\ 2023\]
\[\color {blue}{Solution:}\ \hspace{20cm}\]
\[\text{Given that}\ 75^0\ =\ 30^0\ +\ 45^0\ \text{we use angle sum formula for cosine}\]
\[\hspace{2cm}\ W.\ K.\ T\ cos( A + B )\ =\ cos A\ Cos B\ -\ sin A sin B\]
\[\hspace{3cm}\ cos( 75^0 )\ =\ cos(30^0\ +\ 45^0)\]
\[\hspace{3cm}\ =\ c os\ 30^0\ cos\ 45^0\ -\ sin\ 30^0\ sin\ 45^0\]
\[\hspace{3cm}\ =\ \frac{\sqrt{3}}{2}\ \frac{1}{\sqrt{2}}\ -\ \frac{1}{2}\ \frac{1}{\sqrt{2}}\]
\[\hspace{4cm}\ =\ \frac{\sqrt{3}\ -\ 1}{2\ \sqrt{2}}\]
\[\color {purple} {Example\ 16:}\ \color {red} {Find\ the\ value\ of}\ \frac{Tan\ {25}^0\ +\ Tan\ {20}^0}{1\ -\ Tan\ {25}^0\ Tan\ {20}^0}\ \hspace{15cm}\]
\[\hspace{5cm}\ October\ 2024,\ June(Supp)\ 2025\]
\[\color {blue}{Solution:}\ W.\ K.\ T\ \frac{Tan A\ +\ Tan B}{1\ -\ Tan A\ Tan B}\ =\ Tan(A + B)\ \hspace{18cm}\]
\[\frac{Tan\ {25}^0\ +\ Tan\ {20}^0}{1\ -\ Tan\ {25}^0\ Tan\ {20}^0}\ =\ Tan({25}^0\ +\ {20}^0)\ =\ Tan{45}^0\ =\ 1\ \hspace{10cm}\]
\[\color {purple} {Example\ 17:}\ If\ Tan\ A\ =\ \frac{10}{11} \ and\ Tan\ B\ =\ \frac{1}{21},\ \color {red} {Show\ that\ A\ +\ B\ =\ {45}^0}\ \hspace{15cm}\]
\[\color {blue}{Solution:}\ Given\ Tan\ A\ =\ \frac{10}{11} \ and\ Tan\ B\ =\ \frac{1}{21}\ \hspace{18cm}\]
\[W.\ K.\ T\ Tan(A + B)\ =\ \frac{Tan A\ +\ Tan B}{1\ -\ Tan A\ Tan B}\ \hspace{15cm}\]
\[=\ \frac{\frac{10}{11}\ +\ \frac{1}{21}}{1\ -\ \frac{10}{11}\ ×\ \frac{1}{21}}\ \hspace{10cm}\]
\[=\ \frac{\frac{210\ +\ 11}{231}}{1\ -\ \frac{10}{231}}\ \hspace{10cm}\]
\[=\ \frac{\frac{221}{231}}{\frac{231\ -\ 10}{231}}\ \hspace{10cm}\]
\[=\ \frac{\frac{221}{231}}{\frac{221}{231}}\ \hspace{10cm}\]
\[=\ 1\ \hspace{10cm}\]
\[Tan\ (A\ +\ B)\ =\ 1\ \hspace{10cm}\]
\[\implies\ (A\ +\ B)\ =\ Tan\ ^{-1}\ (1)\ \hspace{10cm}\]
\[\implies\ (A\ +\ B)\ =\ {45}^0\ \hspace{10cm}\]
\[\color {purple} {Example\ 18:}\ If\ A\ and\ B\ are\ acute\ angles\ such\ that\ sin\ A\ =\ \frac{8}{17} \ and\ sin\ B\ =\ \frac{5}{13},\\ \hspace{10cm}\ \color{red}{then\ prove\ that\ sin(A\ +\ B)\ =\ \frac{171}{221}}\ \hspace{8cm}\]
\[\hspace{5cm}\ April\ 2025,\ June(Supp)\ 2025\]
\[\color {blue}{Solution:}\ Given\ Sin\ A\ =\ \frac{8}{17} \ and\ Sin\ B\ =\ \frac{5}{13}\ \hspace{18cm}\]
\[W.\ K.\ T\ Sin ( A + B )\ =\ Sin A\ Cos B\ +\ Cos A\ Sin B\ \hspace{15cm}\]
\[Cos\ A\ =\ ?\ ,\ Cos\ B\ = ?\ \hspace{10cm}\]
\[Cos\ A\ =\ \sqrt{1\ -\ Sin^2\ A}\ \hspace{10cm}\]
\[ =\ \sqrt{1\ -\ (\frac{8}{17})^2}\ \hspace{10cm}\]
\[ =\ \sqrt{1\ -\ \frac{64}{289}}\ \hspace{10cm}\]
\[ =\ \sqrt{\frac{289\ -\ 64}{289}}\ \hspace{10cm}\]
\[ =\ \sqrt{\frac{225}{289}}\ \hspace{10cm}\]
\[Cos\ A\ =\ \frac{15}{17}\ \hspace{10cm}\]
\[Cos\ B\ =\ \sqrt{1\ -\ Sin^2\ B}\ \hspace{10cm}\]
\[ =\ \sqrt{1\ -\ (\frac{5}{13})^2}\ \hspace{10cm}\]
\[ =\ \sqrt{1\ -\ \frac{25}{169}}\ \hspace{10cm}\]
\[ =\ \sqrt{\frac{169\ -\ 25}{169}}\ \hspace{10cm}\]
\[ =\ \sqrt{\frac{144}{169}}\ \hspace{10cm}\]
\[Cos\ B\ =\ \frac{12}{13}\ \hspace{10cm}\]
\[Sin ( A + B )\ =\ (\frac{8}{17})\ (\frac{12}{13})\ +\ (\frac{15}{17})\ (\frac{5}{13})\ \hspace{10cm}\]
\[=\ \frac{96}{221}\ +\ \frac{75}{221}\ \hspace{10cm}\]
\[=\ \frac{96\ +\ 75}{221}\ \hspace{10cm}\]
\[=\ \frac{171}{221}\ \hspace{10cm}\]
\[\boxed {\therefore\ Sin ( A + B )\ =\ \frac{171}{221}}\ \hspace{10cm}\]
\[\color {purple} {Example\ 19:}\ If\ Sin\ A\ =\ \frac{3}{5} \ and\ Cos\ B\ =\ \frac{12}{13},\ \color {red} {Prove\ that\ Sin(A\ +\ B)\ =\ \frac{56}{65}}\ \hspace{15cm}\]
\[\color {blue}{Solution:}\ Given\ Sin\ A\ =\ \frac{3}{5} \ and\ Cos\ B\ =\ \frac{12}{13}\ \hspace{18cm}\]
\[W.\ K.\ T\ Sin ( A + B )\ =\ Sin A\ Cos B\ +\ Cos A\ Sin B\ \hspace{15cm}\]
\[Cos\ A\ =\ ?\ ,\ Sin\ B\ = ?\ \hspace{10cm}\]
\[Cos\ A\ =\ \sqrt{1\ -\ Sin^2\ A}\ \hspace{10cm}\]
\[ =\ \sqrt{1\ -\ (\frac{3}{5})^2}\ \hspace{10cm}\]
\[ =\ \sqrt{1\ -\ \frac{9}{25}}\ \hspace{10cm}\]
\[ =\ \sqrt{\frac{25\ -\ 9}{25}}\ \hspace{10cm}\]
\[ =\ \sqrt{\frac{16}{25}}\ \hspace{10cm}\]
\[Cos\ A\ =\ \frac{4}{5}\ \hspace{10cm}\]
\[Cos\ B\ =\ \sqrt{1\ -\ Sin^2\ B}\ \hspace{10cm}\]
\[ =\ \sqrt{1\ -\ (\frac{12}{13})^2}\ \hspace{10cm}\]
\[ =\ \sqrt{1\ -\ \frac{144}{169}}\ \hspace{10cm}\]
\[ =\ \sqrt{\frac{169\ -\ 144}{169}}\ \hspace{10cm}\]
\[Cos\ B\ =\ \frac{5}{13}\ \hspace{10cm}\]
\[Sin ( A + B )\ =\ (\frac{3}{5})\ (\frac{12}{13})\ +\ (\frac{4}{5})\ (\frac{5}{13})\ \hspace{10cm}\]
\[=\ \frac{36}{25}\ +\ \frac{20}{65}\ \hspace{10cm}\]
\[=\ \frac{36\ +\ 20}{65}\ \hspace{10cm}\]
\[=\ \frac{56}{65}\ \hspace{10cm}\]
\[\boxed {\therefore\ Sin ( A + B )\ =\ \frac{56}{65}}\ \hspace{10cm}\]
\[\color {purple} {Example\ 20:}\ If\ A\ and\ B\ are\ acute\ angles\ and\ Sin\ A\ =\ \frac{1}{\sqrt{10}},\ Sin\ B\ =\ \frac{1}{\sqrt{5}},\ \hspace{15cm}\]\[ \color {red} {Prove\ that\ (A\ +\ B)\ =\ \frac{π}{4}}\ \hspace{10cm}\]
\[\color {blue}{Solution:}\ Given\ Sin\ A\ =\ \frac{1}{\sqrt{10}} \ and\ Sin\ B\ =\ \frac{1}{\sqrt{5}}\ \hspace{18cm}\]
\[W.\ K.\ T\ Sin ( A + B )\ =\ Sin A\ Cos B\ +\ Cos A\ Sin B\ \hspace{15cm}\]
\[Cos\ A\ =\ ?\ ,\ Cos\ B\ = ?\ \hspace{10cm}\]
\[Cos\ A\ =\ \sqrt{1\ -\ Sin^2\ A}\ \hspace{10cm}\]
\[ =\ \sqrt{1\ -\ (\frac{1}{\sqrt{10}})^2}\ \hspace{10cm}\]
\[ =\ \sqrt{1\ -\ \frac{1}{10}}\ \hspace{10cm}\]
\[ =\ \sqrt{\frac{10\ -\ 1}{10}}\ \hspace{10cm}\]
\[ =\ \sqrt{\frac{9}{10}}\ \hspace{10cm}\]
\[Cos\ A\ =\ \frac{3}{\sqrt{10}}\ \hspace{10cm}\]
\[Cos\ B\ =\ \sqrt{1\ -\ Sin^2\ B }\ \hspace{10cm}\]
\[ =\ \sqrt{1\ -\ (\frac{1}{\sqrt{5}})^2}\ \hspace{10cm}\]
\[ =\ \sqrt{1\ -\ \frac{1}{5}}\ \hspace{10cm}\]
\[ =\ \sqrt{\frac{5\ -\ 1}{10}}\ \hspace{10cm}\]
\[ =\ \sqrt{\frac{4}{5}}\ \hspace{10cm}\]
\[Cos\ B\ =\ \frac{2}{\sqrt{5}}\ \hspace{10cm}\]
\[Sin ( A + B )\ =\ (\frac{1}{\sqrt{10}})\ (\frac{2}{\sqrt{5}})\ +\ (\frac{3}{\sqrt{10}})\ (\frac{1}{\sqrt{5}})\ \hspace{10cm}\]
\[=\ \frac{2}{\sqrt{50}}\ +\ \frac{3}{\sqrt{50}}\ \hspace{10cm}\]
\[=\ \frac{2\ +\ 3}{\sqrt{50}}\ \hspace{10cm}\]
\[=\ \frac{5}{\sqrt{50}}\ \hspace{10cm}\]
\[=\ \frac{5}{\sqrt{25 × 2}}\ \hspace{10cm}\]
\[Sin ( A + B )\ =\ \frac{1}{\sqrt{2}}\ \hspace{10cm}\]
\[\boxed {\therefore\ (A\ +\ B)\ =\ \frac{π}{4}}\ \hspace{10cm}\]
\[\color {purple} {Example\ 21:}\ If\ A\ and\ B\ are\ acute\ angles\ such\ that\ Sin\ A\ =\ \frac{3}{5} \ and\ Cos\ B\ =\ \frac{12}{13},\hspace{15cm}\\ \color {red} {then\ prove\ that\ Cos(A\ +\ B)\ =\ \frac{33}{65}}\ \hspace{15cm}\]
\[\hspace{5cm}\ October\ 2024\ June(Supp)\ 2025\]
\[\color {blue}{Solution:}\ Given\ Sin\ A\ =\ \frac{3}{5} \ and\ Cos\ B\ =\ \frac{12}{13}\ \hspace{18cm}\]
\[W.\ K.\ T\ Cos( A + B )\ =\ Cos A\ Cos B\ -\ Sin A\ Sin B\ \hspace{15cm}\]
\[Cos\ A\ =\ ?\ ,\ Sin\ B\ = ?\ \hspace{10cm}\]
\[Cos\ A\ =\ \sqrt{1\ -\ Sin^2\ A}\ \hspace{10cm}\]
\[ =\ \sqrt{1\ -\ (\frac{3}{5})^2}\ \hspace{10cm}\]
\[ =\ \sqrt{1\ -\ \frac{9}{25}}\ \hspace{10cm}\]
\[ =\ \sqrt{\frac{25\ -\ 9}{25}}\ \hspace{10cm}\]
\[ =\ \sqrt{\frac{16}{25}}\ \hspace{10cm}\]
\[Cos\ A\ =\ \frac{4}{5}\ \hspace{10cm}\]
\[Sin\ B\ =\ \sqrt{1\ -\ Cos^2\ B}\ \hspace{10cm}\]
\[ =\ \sqrt{1\ -\ (\frac{12}{13})^2}\ \hspace{10cm}\]
\[ =\ \sqrt{1\ -\ \frac{144}{169}}\ \hspace{10cm}\]
\[ =\ \sqrt{\frac{169\ -\ 144}{169}}\ \hspace{10cm}\]
\[ =\ \sqrt{\frac{25}{169}}\ \hspace{10cm}\]
\[Sin\ B\ =\ \frac{5}{13}\ \hspace{10cm}\]
\[Cos( A + B )\ =\ (\frac{4}{5})\ (\frac{12}{13})\ -\ (\frac{3}{5})\ (\frac{5}{13})\ \hspace{10cm}\]
\[=\ \frac{48}{65}\ -\ \frac{15}{65}\ \hspace{10cm}\]
\[=\ \frac{48\ -\ 15}{65}\ \hspace{10cm}\]
\[=\ \frac{33}{65}\ \hspace{10cm}\]
\[\boxed {Cos ( A + B )\ =\ \frac{33}{65}}\ \hspace{10cm}\]
\[\color {purple} {Example\ 22:}\ If\ Cos\ A\ =\ \frac{1}{7} \ and\ Cos\ B\ =\ \frac{13}{14},\ \color {red} {prove\ that\ A\ -\ B\ =\ \frac{\pi}{3}}\ \hspace{15cm}\]
\[\color {blue}{Solution:}\ Given\ Cos\ A\ =\ \frac{1}{7} \ and\ Cos\ B\ =\ \frac{13}{14}\ \hspace{18cm}\]
\[W.\ K.\ T\ Cos( A – B )\ =\ Cos A\ Cos B\ +\ Sin A\ Sin B\ \hspace{15cm}\]
\[Sin\ A\ =\ ?\ ,\ Sin\ B\ = ?\ \hspace{10cm}\]
\[Sin\ A\ =\ \sqrt{1\ -\ Coś^2\ A}\ \hspace{10cm}\]
\[ =\ \sqrt{1\ -\ (\frac{1}{7})^2}\ \hspace{10cm}\]
\[ =\ \sqrt{1\ -\ \frac{1}{49}}\ \hspace{10cm}\]
\[ =\ \sqrt{\frac{49\ -\ 1}{49}}\ \hspace{10cm}\]
\[ =\ \sqrt{\frac{48}{49}}\ \hspace{10cm}\]
\[ =\ \frac{\sqrt{16\ \times 3}}{\sqrt{49}}\ \hspace{10cm}\]
\[Sin\ A\ =\ \frac{4\ \sqrt{3}}{7}\ \hspace{10cm}\]
\[Sin\ B\ =\ \sqrt{1\ -\ Cos^2\ B}\ \hspace{10cm}\]
\[ =\ \sqrt{1\ -\ (\frac{13}{14})^2}\ \hspace{10cm}\]
\[ =\ \sqrt{1\ -\ \frac{169}{196}}\ \hspace{10cm}\]
\[ =\ \sqrt{\frac{196\ -\ 169}{196}}\ \hspace{10cm}\]
\[ =\ \sqrt{\frac{27}{196}}\ \hspace{10cm}\]
\[ =\ \frac{\sqrt{9\ \times 3}}{\sqrt{196}}\ \hspace{10cm}\]
\[Sin\ B\ =\ \frac{3\ \sqrt{3}}{14}\ \hspace{10cm}\]
\[Cos( A – B )\ =\ ( \frac{1}{7})\ ( \frac{13}{14})\ +\ (\frac{4\ \sqrt{3}}{7})\ (\frac{3\ \sqrt{3}}{14})\ \hspace{10cm}\]
\[=\ \frac{13}{98}\ +\ \frac{36}{98}\ \hspace{10cm}\]
\[=\ \frac{13\ +\ 36}{98}\ \hspace{10cm}\]
\[=\ \frac{49}{98}\ \hspace{10cm}\]
\[Cos ( A – B )\ =\ \frac{1}{2}\ \hspace{10cm}\]
\[\boxed{A\ -\ B\ =\ \frac{\pi}{3}}\ \hspace{10cm}\]
\[\color {purple} {Example\ 23:}\ If\ A\ +\ B\ =\ {45}^0 \ \color {red} {Prove\ that\ (1\ +\ tan\ A)\ (1\ +\ tan\ B)\ =\ 2}\ \hspace{15cm}\]\[\color {red} {Hence\ deduce\ the\ value\ of\ tan\ \ 22\ {\frac{1}{2}}^0}\ \hspace{13cm}\]
\[\hspace{5cm}\ October\ 2023\]
\[\color {blue}{Solution:}\ Given\ A\ +\ B\ =\ 45^{0}\ \hspace{18cm}\]
\[Taking\ Tan\ on\ both\ sides\ \hspace{10cm}\]
\[tan\ (A\ +\ B)\ =\ tan\ 45^{0}\ \hspace{10cm}\]
\[\frac{tan A\ +\ tan B}{1\ -\ tan A\ tan B}\ =\ 1\ \hspace{10cm}\]
\[tan\ A\ +\ tan\ B\ =\ 1\ -\ tan\ A\ tan\ B\ \hspace{10cm}\]
\[tan\ A\ +\ tan\ B\ +\ tan\ A\ tan\ B\ =\ 1\ ————\ (1)\ \hspace{10cm}\]
\[L.\ H.\ S\ =\ (1\ +\ tan\ A)\ (1\ +\ tan\ B)\ \hspace{10cm}\]
\[=\ 1\ +\ tan\ B\ +\ tan\ A\ +\ tan\ A\ tan\ B\ \hspace{10cm}\]
\[=\ 1\ +\ 1\ \hspace{5cm}\ using\ (1)\]
\[=\ 2\ =\ R.\ H.\ S\ \hspace{10cm}\]
\[(1\ +\ tan\ A)\ (1\ +\ tan\ B)\ =\ 2\ \hspace{10cm}\]
\[Put\ B\ =\ A\ \hspace{10cm}\]
\[A\ +\ A\ =\ 45^{0}\ \hspace{10cm}\]
\[2A\ =\ 45^{0}\ \hspace{10cm}\]
\[A\ =\ 22\ {\frac{1}{2}}^0\ \hspace{10cm}\]
\[(1\ +\ tan\ 22\ {\frac{1}{2}}^0)\ (1\ +\ Tan\ 22\ {\frac{1}{2}}^0)\ =\ 2\ \hspace{10cm}\]
\[(1\ +\ tan\ 22\ {\frac{1}{2}}^0)^2\ =\ 2\ \hspace{10cm}\]
\[1\ +\ tan\ 22\ {\frac{1}{2}}^0\ =\ \sqrt{2}\ \hspace{10cm}\]
\[\boxed {tan\ 22\ {\frac{1}{2}}^0\ =\ \sqrt{2}\ -\ 1}\hspace{10cm}\]
\[\color {royalblue} {Multiple\ Angles\ of\ 2A}:\ \hspace{20cm}\]
\[\color {brown} {Formulae}:\ \hspace{20cm}\]
\[1\ (i)\ Sin\ 2A\ =\ 2\ Sin\ A\ Cos\ A\ \hspace{5cm}\ (ii)\ Sin\ 2A\ =\ \frac{2\ Tan\ A}{1\ +\ Tan^2\ A}\]
\[2\ (i)\ Cos\ 2A\ =\ Cos^2A\ -\ Sin^2A\ \hspace{5cm}\ (ii)\ Cos\ 2A\ =\ \frac{1\ -\ Tan^2A}{1\ +\ Tan^2\ A}\]
\[3.\ Tan\ 2A\ =\ \frac{2\ Tan\ A}{1\ -\ Tan^2\ A}\ \hspace{10cm}\]
\[4.\ Sin^2A\ =\ \frac{1\ -\ Cos\ 2A}{2}\ \hspace{5cm}\ Note:\ 1\ -\ 2\ Sin^2A\ =\ Cos\ 2A\]
\[5.\ Cos^2A\ =\ \frac{1\ +\ Cos\ 2A}{2}\ \hspace{10cm}\]
\[6.\ Tan^2A\ =\ \frac{1\ -\ Cos\ 2A}{1\ +\ Cos\ 2A}\ \hspace{10cm}\]
\[\color {purple} {Example\ 24:}\ \color {red} {Prove\ that}\ \frac{sin\ 2A}{1\ +\ cos\ 2A}\ =\ tan\ A.\ \hspace{15cm}\]
\[\hspace{5cm}\ April\ 2024\]
\[\color {blue}{Solution:}\ LHS\ =\ \frac{sin\ 2A}{1\ +\ cos\ 2A}\ \hspace{18cm}\]
\[=\ \frac{2\ sin\ A\ cos\ A}{2\ cos^2\ A}\ \hspace{2cm}\ W.\ K.\ T.\ sin\ 2A\ =\ 2\ sin\ A\ cos\ A\ and\ 1\ +\ cos\ 2A\ =\ 2\ cos^2\ A\]
\[=\ \frac{sin\ A}{cos\ A}\ \hspace{15cm}\]
\[=\ tan\ A\ =\ R.H.S\ \hspace{15cm}\]
\[\color {purple} {Example\ 25:}\ \color {red} {Find\ the\ value\ of}\ 2\ Sin\ 30^{0}\ Cos\ 30^{0}\ \hspace{15cm}\]
\[\color {blue}{Solution:}\ 2\ Sin\ 30^{0}\ Cos\ 30^{0}\ =\ Sin\ 2(30^{0})\ \hspace{15cm}\]
\[=\ Sin\ 60^{0}\ \hspace{15cm}\]
\[=\ \frac{\sqrt{3}}{2}\ \hspace{15cm}\]
\[\boxed {2\ Sin\ 30^{0}\ Cos\ 30^{0}\ =\ \frac{\sqrt{3}}{2}}\]
\[\color {purple} {Example\ 26:}\ \color {red} {Find\ the\ value\ of}\ 2\ Sin\ 15^{0}\ Cos\ 15^{0}\ \hspace{15cm}\]
\[\hspace{5cm}\ October\ 2024,\ June(Supp)\ 2025\]
\[\color {blue}{Solution:}\ 2\ Sin\ 15^{0}\ Cos\ 15^{0}\ =\ Sin\ 2(15^{0})\ \hspace{15cm}\]
\[=\ Sin\ 30^{0}\ \hspace{15cm}\]
\[=\ \frac{1}{2}\ \hspace{15cm}\]
\[\boxed {2\ Sin\ 30^{0}\ Cos\ 30^{0}\ =\ \frac{1}{2}}\]
\[\color {purple} {Example\ 27:}\ \color {red} {Find\ the\ value\ of}\ Cos^2\ 15^{0}\ -\ Sin^2\ 15^{0}\ \hspace{15cm}\]
\[\color {blue}{Solution:}\ Cos^2\ 15^{0}\ -\ Sin^2\ 15^{0}\ =\ \hspace{18cm}\]
\[=\ Cos\ 2(15^{0})\ \hspace{15cm}\]
\[=\ Cos\ 30^{0}\ \hspace{15cm}\]
\[=\ \frac{\sqrt{3}}{2}\ \hspace{15cm}\]
\[\boxed {Cos^2\ 15^{0}\ -\ Sin^2\ 15^{0}\ =\ \frac{\sqrt{3}}{2}}\]
\[\color {purple} {Example\ 28:}\ \color {red} {Find\ the\ value\ of}\ 1\ -\ 2\ Sin^2\ 45^{0}\ \hspace{15cm}\]
\[\color {blue}{Solution:}\ 1\ -\ 2\ Sin^2\ 45^{0}\ =\ \hspace{18cm}\]
\[=\ Cos\ 2(45^{0})\ \hspace{15cm}\]
\[=\ Cos\ 90^{0}\ \hspace{15cm}\]
\[=\ 0\ \hspace{15cm}\]
\[\color {purple} {Example\ 29:}\ \color {red} {Show\ that}\ \frac{sin\ A\ +\ sin\ 2A}{1\ +\ cos\ A\ +\ cos\ 2A}\ =\ tan\ A\ \hspace{15cm}\]
\[\hspace{5cm}\ October\ 2023,\ April\ 2024,\ April\ 2025\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\hspace{2cm}\ LHS\ =\ \frac{sin\ A\ +\ sin\ 2A}{1\ +\ Cos\ A\ +\ cos\ 2A}\ \hspace{18cm}\]
\[=\ \frac{sin\ A\ +\ 2\ sin\ A\ Cos\ A}{2\ cos^2\ A\ +\ Cos\ A}\ \hspace{2cm}\ W.\ K.\ T.\ Sin\ 2A\ =\ 2\ Sin\ A\ Cos\ A\ , and\ 1\ +\ cos\ 2A\ =\ 2\ cos^2\ A\]
\[=\ \frac{Sin\ A(1\ +\ 2\ Cos\ A)}{Cos\ A(1\ +\ 2\ Cos\ A)}\ \hspace{15cm}\]
\[=\ \frac{Sin\ A}{CosS\ A}\ \hspace{15cm}\]
\[=\ tan\ A\ =\ R.H.S\ \hspace{15cm}\]
\[\color {purple} {Example\ 30:}\ \color {red} {Show\ that}\ \frac{1\ -\ Cos\ 2A\ +\ Sin\ 2A}{1\ +\ Cos\ 2A\ +\ Sin\ 2A}\ =\ Tan\ A\ \hspace{15cm}\]
\[\color {blue}{Solution:}\ LHS\ =\ \frac{1\ -\ Cos\ 2A\ +\ Sin\ 2A}{1\ +\ Cos\ 2A\ +\ Sin\ 2A}\ \hspace{18cm}\]
\[=\ \frac{2\ Sin^2\ A\ +\ 2\ Sin\ A\ Cos\ A}{2\ Cos^2\ A\ +\ 2\ Sin\ A\ Cos\ A}\ \hspace{2cm}\ W.\ K.\ T.\ Sin\ 2A\ =\ 2\ Sin\ A\ Cos\ A\ ,\ 1\ -\ Cos\ 2A\ =\ 2\ Sin^2\ A\ and\ 1\ +\ Cos\ 2A\ =\ 2\ Cos^2\ A\]
\[=\ \frac{2\ Sin\ A(Sin\ A\ +\ Cos\ A)}{2\ Cos\ A(Sin\ A\ +\ Cos\ A}\ \hspace{15cm}\]
\[=\ \frac{Sin\ A}{CosS\ A}\ \hspace{15cm}\]
\[=\ tan\ A\ =\ R.H.S\ \hspace{15cm}\]
\[\color {purple} {Example\ 31:}\ \color {red} {Show\ that}\ \frac{1\ +\ cos\ 2A\ +\ sin\ 2A}{1\ -\ cos\ 2A\ +\ sin\ 2A}\ =\ cot\ A\ \hspace{15cm}\]
\[\hspace{5cm}\ June(Supp)\ 2025\]
\[\color {blue}{Solution:}\ LHS\ =\ \frac{1\ +\ cos\ 2A\ +\ sin\ 2A}{1\ -\ cos\ 2A\ +\ sin\ 2A}\ \hspace{18cm}\]
\[=\ \frac{2\ cos^2\ A\ +\ 2\ sin\ A\ cos\ A}{2\ sin^2\ A\ +\ 2\ sin\ A\ cos\ A}\ \hspace{1cm}\ W.\ K.\ T.\ sin\ 2A\ =\ 2\ sin\ A\ cos\ A\ ,\ 1\ +\ cos\ 2A\ =\ 2\ cos^2\ A\ and\ 1\ -\ cos\ 2A\ =\ 2\ sin^2\ A\]
\[=\ \frac{2\ cos\ A(cos\ A\ +\ sin\ A)}{2\ sin\ A(sin\ A\ +\ cos\ A)}\ \hspace{15cm}\]
\[=\ \frac{con\ A}{sin\ A}\ \hspace{15cm}\]
\[=\ cot\ A\ =\ R.H.S\ \hspace{15cm}\]
\[\color {purple} {Example\ 32:}\ \color {red} {Prove\ that}\ Cos^4\ A\ -\ Sin^4\ A\ =\ Cos\ 2A\ \hspace{15cm}\]
\[\color {blue}{Solution:}\ L.\ H.\ S\ =\ Cos^4\ A\ -\ Sin^4\ A\ \hspace{18cm}\]
\[=\ (Cos^2\ A)^2\ -\ (Sin^2\ A)^2\hspace{15cm}\]
\[=\ (Cos^2\ A\ +\ Sin^2\ A)\ (Cos^2\ A\ +\ Sin^2\ A)\hspace{10cm}\]
\[=\ (1)\ (Cos^2\ A\ -\ Sin^2\ A)\hspace{10cm}\]
\[=\ Cos\ 2A\ =\ R.H.S\ \hspace{15cm}\]
\[\color {purple} {Example\ 33:}\ \color {red} {Prove\ that}\ Sin^2\ A\ +\ Sin^2\ (60^0\ +\ A)\ +\ Sin^2\ (60^0\ -\ A)\ =\ \frac{3}{2}\ \hspace{15cm}\]
\[\color {blue}{Solution:}\ L.\ H.\ S\ =\ Sin^2\ A\ +\ Sin^2\ (60^0\ +\ A)\ +\ Sin^2\ (60^0\ -\ A)\ \hspace{16cm}\]
\[=\ \frac{1\ -\ Cos\ 2A}{2}\ +\ \frac{1\ -\ Cos\ 2(60^0\ +\ A)}{2}\ +\ \frac{1\ -\ Cos\ 2(60^0\ -\ A)}{2}\ \hspace{10cm}\]
\[=\ \frac{1}{2}\ +\ \frac{1}{2}\ +\ \frac{1}{2}\ -\ \frac{1}{2}[Cos\ 2\ A\ +\ Cos\ (120^0\ +\ 2\ A)\ +\ Cos\ (120^0\ -\ 2\ A)]\ \hspace{10cm}\]
\[=\ \frac{3}{2}\ -\ \frac{1}{2}[Cos\ 2\ A\ +\ 2\ Cos\ 120^0\ Cos\ 2\ A]\ \hspace{2cm}\ \because\ Cos(A\ +\ B)\ +\ Cos(A\ -\ B)\ =\ 2\ Cos\ A\ Cos\ B\]
\[=\ \frac{3}{2}\ -\ \frac{1}{2}[Cos\ 2\ A\ +\ 2\ (\frac{-1}{2})\ Cos\ 2\ A]\ \hspace{10cm}\]
\[=\ \frac{3}{2}\ -\ \frac{1}{2}[Cos\ 2\ A\ -\ Cos\ 2\ A]\ \hspace{10cm}\]
\[=\ \frac{3}{2}\ -\ \frac{1}{2}[0]\ \hspace{10cm}\]
\[=\ \frac{3}{2}\ =\ R.H.S\ \hspace{10cm}\]
\[\color {purple} {Example\ 34:}\ If\ tan\ A\ =\ \frac{1}{2},\ \color {red} {find\ tan\ 2A}\ \hspace{15cm}\]
\[\hspace{5cm}\ October\ 2023\ April\ 2025\]
\[\color {blue}{Solution:}\ \hspace{20cm}\]
\[\hspace{2cm}\ Given\ tan\ A\ =\ \frac{1}{2}\ \hspace{18cm}\]
\[\hspace{2cm}\ W.\ K.\ T\ tan\ 2A\ =\ \frac{2\ tan\ A}{1\ -\ tan^2\ A}\ \hspace{10cm}\]
\[\hspace{3cm}\ =\ \frac{2\ \times\ \frac{1}{2}}{1\ -\ (\frac{1}{2})^2}\ \hspace{10cm}\]
\[\hspace{3cm}\ =\ \frac{1}{1\ -\ \frac{1}{4}}\ \hspace{10cm}\]
\[\hspace{3cm}\ =\ \frac{1}{\frac{4\ -\ 1}{4}}\ \hspace{10cm}\]
\[\hspace{3cm}\ =\ \frac{1}{\frac{3}{4}}\ \hspace{10cm}\]
\[\hspace{4cm}\ =\ \frac{4}{3}\ \hspace{10cm}\]
\[\color {purple} {Example\ 35:}\ If\ Tan\ A\ =\ \frac{1}{3} \ and\ Tan\ B\ =\ \frac{1}{7},\ \color {red} {Show\ that\ 2A\ +\ B\ =\ \frac{π}{4}}\ \hspace{15cm}\]
\[\color {blue}{Solution:}\ Given\ Tan\ A\ =\ \frac{1}{3} \ and\ Tan\ B\ =\ \frac{1}{7}\ \hspace{18cm}\]
\[Tan(A + B)\ =\ \frac{Tan A\ +\ Tan B}{1\ -\ Tan A\ Tan B}\ \hspace{15cm}\]
\[Tan(2A + B)\ =\ \frac{Tan 2A\ +\ Tan B}{1\ -\ Tan 2A\ Tan B}\ —————- (1)\ \hspace{15cm}\]
\[Tan\ 2A\ =\ \frac{2\ Tan\ A}{1\ -\ Tan^2\ A}\ \hspace{10cm}\]
\[ =\ \frac{2\ (\frac{1}{3})}{1\ -\ (\frac{1}{3})^2}\ \hspace{10cm}\]
\[ =\ \frac{\frac{2}{3}}{1\ -\ \frac{1}{9}}\ \hspace{10cm}\]
\[ =\ \frac{\frac{2}{3}}{\frac{8}{9}}\ \hspace{10cm}\]
\[ =\ \frac{2}{3}\ ×\ \frac{9}{8}\ \hspace{10cm}\]
\[ =\ \frac{3}{4}\ \hspace{10cm}\]
\[Equation\ (1)\ becomes\]
\[Tan(2A + B)\ =\ \frac{\frac{3}{4}\ +\ \frac{1}{7}}{1\ -\ \frac{3}{4}\ ×\ \frac{1}{7}}\ \hspace{15cm}\]
\[=\ \frac{\frac{21\ +\ 4}{28}}{1\ -\ \frac{3}{28}}\ \hspace{10cm}\]
\[=\ \frac{\frac{25}{28}}{\frac{28\ -\ 3}{28}}\ \hspace{10cm}\]
\[=\ \frac{\frac{25}{28}}{\frac{25}{28}}\ \hspace{10cm}\]
\[=\ 1\ \hspace{10cm}\]
\[Tan\ (2A\ +\ B)\ =\ 1\ \hspace{10cm}\]
\[\implies\ (2A\ +\ B)\ =\ Tan\ ^{-1}\ (1)\ \hspace{10cm}\]
\[\boxed{\therefore\ (2A\ +\ B)\ =\ \frac{π}{4}}\ \hspace{10cm}\]
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