# SOLUTIONS TO ASSIGNMENT-2 FOR ENGINEERING MATHEMATICS – I

$1.\ Prove\ that\ \frac{1\ -\ Cos\ 2A\ +\ Sin\ 2A}{1\ +\ Cos\ 2A\ +\ Sin\ 2A}\ =\ Tan\ A\ \hspace{15cm}$
$\color {black}{Solution:}\ LHS\ =\ \frac{1\ -\ Cos\ 2A\ +\ Sin\ 2A}{1\ +\ Cos\ 2A\ +\ Sin\ 2A}\ \hspace{18cm}$
$W.\ K.\ T.\ Sin\ 2A\ =\ 2\ Sin\ A\ Cos\ A\ ,\ 1\ +\ Cos\ 2A\ =\ 2\ Cos^2\ A\ and\ 1\ +\ Cos\ 2A\ =\ 2\ Cos^2\ A$
$\color {black}{Solution:}\ LHS\ =\ \frac{2\ Sin^2\ A\ +\ 2\ Sin\ A\ Cos\ A}{2\ Cos^2\ A\ +\ 2\ Sin\ A\ Cos\ A}\ \hspace{18cm}$
$=\ \frac{2Sin\ A{Sin\ A\ +\ Cos\ A}{2Cos\ A(Sin\ A\ +\ Cos\ A}\ \hspace{15cm}$
$=\ \frac{Sin\ A}{Cos\ A}\ \hspace{15cm}$
$=\ Tan\ A\ =\ R.H.S\ \hspace{15cm}$
$2.\ Prove\ that\ Cos\ 10^{0}\ Cos\ 50^{0}\ Cos\ 70^{0}\ =\ \frac{\sqrt{3}}{8}\ \hspace{15cm}$
$\color {black}{Solution:}\ \hspace{19cm}$
$W.\ K.\ T\ Cos\ (A\ +\ B)Cos\ (A\ -\ B)\ =\ Cos^2\ A\ -\ Sin^2\ B\ and$
$Cos\ 3A\ =\ 4\ Cos^3\ A\ -\ 3\ Cos\ A$
$L.\ H.\ S\ =\ Cos\ 10^{0}\ Cos\ 50^{0}\ Cos\ 70^{0}\ \hspace{10cm}$
$=\ Cos\ 10^{0}\ Cos\ (60^{0}\ -\ 10^{0}) Cos\ (60^{0}\ +\ 10^{0})\ \hspace{10cm}$
$=\ Cos\ 10^{0}\ (Cos^2\ 60^{0}\ -\ Sin^2\ 10^{0})\ \hspace{10cm}$
$=\ Cos\ 10^{0}\ (\frac{1}{4}\ -\ (1\ -\ Cos^2\ 10^{0}))\ \hspace{10cm}$
$=\ Cos\ 10^{0}\ [\frac{1\ -\ 4\ +\ 4\ Cos^2\ 10^{0}}{4}]\ \hspace{10cm}$
$=\ \frac{1}{4}\ Cos\ 10^{0}\ [-\ 3\ +\ 4\ Cos^2\ 10^{0}]\ \hspace{10cm}$
$=\ \frac{1}{4}(4\ Cos^3\ 10^{0}\ -\ 3\ Cos\ 10^{0})\ \hspace{10cm}$
$=\ \frac{1}{4}\ Cos\ 3(10^{0})\ \hspace{10cm}$
$=\ \frac{1}{4}\ Cos\ 30^{0}\ \hspace{10cm}$
$=\ \frac{1}{4}\ (\frac{\sqrt{3}}{2})\ \hspace{10cm}$
$=\ \frac{\sqrt{3}}{8}\ =\ R.H.S\ \hspace{10cm}$
$3.\ Evaluate:\ \lim\ _{x\ \to\ 2}\ \frac{x^3\ -\ 8}{x^4\ -\ 16}\ \hspace{15cm}$
$\color {black}{Solution:}\ W.\ K.\ T\ \lim\ _{x\ \to\ a}\ \frac{x^n\ -\ a^n}{x\ -\ a}\ =\ n\ a^{n\ -\ 1}\ \hspace{15cm}$
$\lim\ _{x\ \to\ 2}\ \frac{x^3\ -\ 8}{x^4\ -\ 16}\ =\ \lim\ _{x\ \to\ 2}\ \frac{x^3\ -\ 2^3}{x^4\ -\ 2^4} \hspace{10cm}$
$Multiply\ and\ divide\ by\ ‘x – 2$
$\lim\ _{x\ \to\ 2}\ \frac{\frac{x^5\ -\ 2^5}{x\ -\ 2}}{\frac{x^3\ -\ 2^3}{x\ -\ 2}}\ \hspace{10cm}$
$=\ \frac{5\ 2^{5\ -\ 1}}{3\ 2^{3\ -\ 1}}\ \hspace{10cm}$
$=\ \frac{5\ 2^4}{3\ 2^2}\ \hspace{10cm}$
$=\ \frac{5\ 2^{4\ -\ 2}}{3}\ \hspace{10cm}$
$=\ \frac{5\ (2^2)}{3}\ \hspace{10cm}$
$=\ \frac{5\ (4)}{3}\ \hspace{10cm}$
$=\ \frac{20}{3}\ \hspace{10cm}$