Question

prove that :-
$\frac{ {tan}^{2} \alpha }{sec \alpha + 1} = \frac{ {1 - cos \alpha } }{cos \alpha }$
​

Answer 1

Step-by-step explanation:

$rhs \: \: = \frac{ {tan}^{2} \alpha }{ \sec\alpha + 1 }$

we know that ,

${tan}^{2} \alpha = {sec}^{2} \alpha - 1$

$=( { \sec\alpha })^{2} - ({1})^{2}$

$= ( \sec\alpha - 1)(sec \alpha + 1)$

So ,

Substitute this for tan^2alpha

$rhs \: \: = \: \: \frac{(sec \alpha + 1)(sec \alpha - 1)}{(sec \alpha + 1)}$

$= sec \alpha - 1$

$= \frac{1}{cos \alpha } - 1$

$= \frac{1 - cos \alpha }{cos \alpha } = rhs$

Hence proved ! !

Answer 2

SOLUTION ☺️

$= > \frac{ \tan {}^{2} \alpha }{ \sec \alpha + 1} = \frac{1 - cos \alpha }{ \cos \alpha } \\ \\ = > tan {}^{2} \alpha = sec {}^{2} \alpha + 1 \\ = > (sec \ \alpha ) {}^{2} + (1) {}^{2} \\ = > (sec \: \alpha + 1) \: ( \sec \alpha - 1) \\ = > substitute \: the \: place \: of \: tan \: {}^{2} \alpha \\ = > \frac{(sec \alpha + 1)(sec \alpha - 1)}{( \sec \alpha + 1) } \\ \\ = > \sec \alpha - 1 \\ = > 1 - \frac{1}{cos \alpha } \\ \\ = > \frac{1 - cos \alpha }{ \cos \alpha }$

hope it helps ✔️