# SOLUTIONS TO SEMINAR – 2 FOR ENGINEERING MATHEMATICS – I

$Group\ 1:\ If\ Sin\ A\ =\ \frac{8}{17} \ and\ Sin\ B\ =\ \frac{5}{13},\ Show\ that\ Sin(A\ +\ B)\ =\ \frac{171}{221}\ \hspace{15cm}$
$\color {black}{Solution:}\ Given\ Sin\ A\ =\ \frac{8}{17} \ and\ Cos\ B\ =\ \frac{5}{13}\ \hspace{18cm}$
$W.\ K.\ T\ Sin ( A + B )\ =\ Sin A\ Cos B\ +\ Cos A\ Sin B\ \hspace{15cm}$
$Cos\ A\ =\ ?\ ,\ Cos\ B\ = ?\ \hspace{10cm}$
$Cos\ A\ =\ \sqrt{1\ -\ Sin^2\ A}\ \hspace{10cm}$
$=\ \sqrt{1\ -\ (\frac{8}{17})^2}\ \hspace{10cm}$
$=\ \sqrt{1\ -\ \frac{64}{289}}\ \hspace{10cm}$
$=\ \sqrt{\frac{289\ -\ 64}{289}}\ \hspace{10cm}$
$=\ \sqrt{\frac{225}{289}}\ \hspace{10cm}$
$Cos\ A\ =\ \frac{15}{17}\ \hspace{10cm}$
$Cos\ B\ =\ \sqrt{1\ -\ Sin^2\ B}\ \hspace{10cm}$
$=\ \sqrt{1\ -\ (\frac{5}{13})^2}\ \hspace{10cm}$
$=\ \sqrt{1\ -\ \frac{25}{169}}\ \hspace{10cm}$
$=\ \sqrt{\frac{169\ -\ 25}{169}}\ \hspace{10cm}$
$=\ \sqrt{\frac{25}{144}}\ \hspace{10cm}$
$Cos\ B\ =\ \frac{5}{12}\ \hspace{10cm}$
$Sin ( A + B )\ =\ (\frac{8}{17})\ (\frac{12}{13})\ +\ (\frac{15}{17})\ (\frac{5}{13})\ \hspace{10cm}$
$=\ \frac{96}{221}\ +\ \frac{75}{221}\ \hspace{10cm}$
$=\ \frac{96\ +\ 75}{221}\ \hspace{10cm}$
$=\ \frac{171}{221}\ \hspace{10cm}$
$Group\ 2:\ If\ Tan\ A\ =\ \frac{1}{2} \ and\ Tan\ B\ =\ \frac{1}{3},\ find\ the\ value\ of\ tan(A\ +\ B)\ \hspace{15cm}$
$Solution:\ W.\ K.\ T\ Tan(A + B)\ =\ \frac{Tan A\ +\ Tan B}{1\ -\ Tan A\ Tan B}\ \hspace{15cm}$
$=\ \frac{\frac{1}{2}\ +\ \frac{1}{3}}{1\ -\ \frac{1}{2}\ ×\ \frac{1}{3}}\ \hspace{10cm}$
$=\ \frac{\frac{3\ +\ 2}{6}}{1\ -\ \frac{1}{6}}\ \hspace{10cm}$
$=\ \frac{\frac{5}{6}}{\frac{6\ -\ 1}{6}}\ \hspace{10cm}$
$=\ \frac{\frac{5}{6}}{\frac{5}{6}}\ \hspace{10cm}$
$=\ 1\ \hspace{10cm}$
$Tan\ (A\ +\ B)\ =\ 1\ \hspace{10cm}$
$Group\ 3:\ If\ Cos\ θ\ =\ \frac{1}{3},\ find\ Cos\ 3θ\ \hspace{15cm}$
$\color {black}{Solution:}\ Given\ Cos\ θ\ =\ \frac{1}{3}\ \hspace{18cm}$
$W.\ K.\ T\ Cos\ 3\ \theta\ =\ 4\ Cos^3\ \theta\ -\ 3\ Cos\ \theta\ – \hspace{10cm}$
$=\ 4(\frac{1}{3})^3\ -\ 3(\frac{1}{3})\ \hspace{10cm}$
$=\ \frac{4}{27}\ -\ 1\ \hspace{10cm}$
$=\ \frac{4\ -\ 27}{27}\ \hspace{10cm}$
$=\ \frac{-23}{27}\ \hspace{10cm}$
$Group\ 4.\ Prove\ that\ \frac{Sin\ 3\ \theta}{Sin\ \theta}\ -\ \frac{Cos\ 3\ \theta}{Cos\ \theta}\ =\ 2\ \hspace{15cm}$
$\color {black}{Solution:}\ L.\ H.\ S\ =\ \frac{Sin\ 3\ \theta}{Sin\ \theta}\ -\ \frac{Cos\ 3\ \theta}{Cos\ \theta}\ \hspace{18cm}$
$=\ \frac{3\ Sin\ \theta\ -\ 4\ Sin^3\ \theta}{Sin\ \theta}\ -\ \frac{4\ Cos^3\ \theta\ -\ 3\ Cos\ \theta}{Cos\ \theta}\ \hspace{15cm}$
$=\ \frac{Sin\ \theta(3\ -\ 4\ Sin^2\ \theta)}{Sin\ \theta}\ -\ \frac{Cos\ \theta(4\ Cos^2\ \theta\ -\ 3)}{Cos\ \theta}\ \hspace{10cm}$
$=\ 3\ -\ 4\ Sin^2\ \theta\ -\ (4\ Cos^2\ \theta\ -\ 3)\ \hspace{10cm}$
$=\ 3\ -\ 4\ Sin^2\ \theta\ -\ 4\ Cos^2\ \theta\ +\ 3\ \hspace{10cm}$
$=\ 6\ -\ 4(Sin^2\ \theta\ +\ Cos^2\ \theta)\ \hspace{10cm}$
$=\ 6\ -\ 4(1)\ \hspace{10cm}$
$=\ 6\ -\ 4\ \hspace{10cm}$
$\frac{Sin\ 3\ \theta}{Sin\ \theta}\ -\ \frac{Cos\ 3\ \theta}{Cos\ \theta}\ =\ 2\ \hspace{10cm}$
$Group5.\ If\ a\ =\ Sin\ x\ +\ Sin\ y,\ b\ =\ Cos\ x\ +\ Cos\ y,\ Prove\ that\ a^2\ +\ b^2\ =\ 4\ Cos^2\ (\frac{x\ -\ y}{2})\ \hspace{15cm}$
$\color {black}{Solution:}\ \hspace{19cm}$
$W.\ K.\ T\ Sin\ A\ +\ Sin\ B =\ 2\ Sin\ (\frac{A\ +\ B}{2})\ Cos\ (\frac{A\ -\ B}{2})\ and$
$Cos\ A\ +\ Cos\ B =\ 2\ Cos\ (\frac{A\ +\ B}{2})\ Cos\ (\frac{A\ -\ B}{2})$
$L.\ H.\ S\ =\ a^2\ +\ b^2\ \hspace{10cm}$
$=\ (Sin\ x\ +\ Sin\ y)^2\ +\ (Cos\ x\ +\ Cos\ y)^2\ \hspace{10cm}$
$=\ [2\ Sin\ (\frac{x\ +\ y}{2})\ Cos\ (\frac{x\ -\ y}{2})]^2\ +\ [2\ Cos\ (\frac{x\ +\ y}{2})\ Cos\ (\frac{x\ -\ y}{2})]^2\ \hspace{10cm}$
$=\ 4\ Sin^2\ (\frac{x\ +\ y}{2})\ Cos^2\ (\frac{x\ -\ y}{2})\ +\ 4\ Cos^2\ (\frac{x\ +\ y}{2})\ Cos^2\ (\frac{x\ -\ y}{2}) \hspace{10cm}$
$=\ 4\ Cos^2\ (\frac{x\ -\ y}{2})[Sin^2\ (\frac{x\ +\ y}{2})\ +\ Cos^2\ (\frac{x\ +\ y}{2})]\ \hspace{10cm}$
$=\ 4\ Cos^2\ (\frac{x\ -\ y}{2})\ \hspace{10cm}$
$Group\ 6:\ Show\ that\ Tan^{-1}\ (\frac{x\ -\ y}{1\ +\ x\ y})\ =\ Tan^{-1}\ x\ – Tan^{-1}\ y\ \hspace{15cm}$
$\color {black}{Solution:}\ \hspace{19cm}$
$Let\ Tan^{-1}\ (x)\ =\ A,\ \implies\ x\ =\ Tan\ A\ \hspace{10cm}$
$Let\ Tan^{-1}\ (y)\ =\ B,\ \implies\ y\ =\ Tan\ B\ \hspace{10cm}$
$W.\ K.\ T\ Tan(A – B)\ =\ \frac{Tan A\ -\ Tan B}{1\ +\ Tan A\ Tan B}\ \hspace{10cm}$
$Tan(A – B)\ =\ \frac{x\ -\ y}{1\ +\ x\ y}\ \hspace{10cm}$
$A – B\ =\ Tan^{-1}(\frac{x\ -\ y}{1\ +\ x\ y})\ \hspace{10cm}$
$Tan^{-1}\ x\ – Tan^{-1}\ y\ =\ Tan^{-1}\ (\frac{x\ -\ y}{1\ +\ x\ y})\ \hspace{10cm}$
$Group\ 7:\ Evaluate:\ \lim\ _{x\ \to\ 3}\ \frac{x^5\ -\ 3^5}{x^4\ -\ 3^4}\ \hspace{15cm}$
$\color {black}{Solution:}\ W.\ K.\ T\ \lim\ _{x\ \to\ a}\ \frac{x^n\ -\ a^n}{x\ -\ a}\ =\ n\ a^{n\ -\ 1}\ \hspace{15cm}$
$Multiply\ and\ divide\ by\ ‘x – 3’$
$=\ \lim\ _{x\ \to\ 3}\ \frac{\frac{x^5\ -\ 3^5}{x\ -\ 3}}{\frac{x^4\ -\ 3^4}{x\ -\ 3}}\ \hspace{10cm}$
$=\ \frac{5\ 3^{5\ -\ 1}}{4\ 3^{4\ -\ 1}}\ \hspace{10cm}$
$=\ \frac{5\ 3^4}{4\ (3^3)}\ \hspace{10cm}$
$=\ \frac{5\ (3^{4\ -\ 3})}{4}\ \hspace{10cm}$
$=\ \frac{5\ (3)}{4}\ \hspace{10cm}$
$=\ \frac{15}{4}\ \hspace{10cm}$