Question

Sin theta +cos theta /sin theta - cos theta +sin theta - cos theta /sin theta +cos theta=2 sec squre theta /tan squre theta -1

Answer 1

Solution:

$\frac{sin \: \theta + cos \: \theta}{sin \: \theta - cos \: \theta} + \frac{sin \: \theta - cos \: \theta}{sin \: \theta + cos \: \theta} \\ \\ \frac{ {(sin \: \theta + cos \: \theta)}^{2} +{(sin \: \theta - cos \: \theta)}^{2} }{ { {sin}^{2} \theta - {cos}^{2}\theta } } \\ \\ \frac{{sin}^{2} \theta + {cos}^{2} \theta + 2sin \: \theta \: cos \: \theta +{sin}^{2} \theta + {cos}^{2} \theta - 2sin \: \theta \: cos \: \theta }{{sin}^{2} \theta - {cos}^{2}\theta}$
Since

${sin}^{2} \theta + {cos}^{2} \theta = 1$
So

$\frac{1 + 1}{ {sin}^{2}\theta - {cos}^{2}\theta } \\ \\ = \frac{2}{ {cos}^{2}\theta\bigg( \frac{ {sin}^{2}\theta }{ {cos}^{2}\theta } - 1\bigg) }$
As

$\frac{1}{cos \: \theta} = sec \: \theta \\ \\ \frac{sin \: \theta}{cos \:\theta} \: = tan \: \theta$
So

$= \frac{2 {sec}^{2}\theta }{ {tan}^{2}\theta - 1 } \\ \\ hence \: proved$

Hope it helps you.

Answer 2

Answer:

Hera is your answer hope it helps........ ✌️