Question

Sin theta minus cos theta by sin theta + cos theta + sin theta + cos theta whole divided by sin theta minus cos theta is equal to 2 by 2 sin squared theta minus one

Answer 1

Step-by-step explanation:

your answer attached in the photo

Answer 2

Given:

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

To Prove:

L.H.S= R.H.S

Solution:

Formula:

$\bold {(a+b)^2=a^2+b^2+2ab}\\\\\bold {(a-b)^2=a^2-b^2-2ab}\\\\\bold {\sin^2 \theta+ \cos^2 \theta= 1}\\\\\bold {\cos^2 \theta =1- \sin^2 \theta}$

Solve equation:

$\Rightarrow \bold{\frac{\sin \theta -\cos \theta}{\sin \theta + \cos \theta} + \frac{\sin \theta +\cos \theta}{\sin \theta - \cos \theta} \ = \frac{2}{2\sin^2 \theta-1}}$

Solve L.H.S.

$\Rightarrow \frac{\sin \theta -\cos \theta}{\sin \theta + \cos \theta} + \frac{\sin \theta +\cos \theta}{\sin \theta - \cos \theta}$

taking L.C.M.

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

$\Rightarrow \frac{\sin^2 \theta +\cos^2 \theta -2\sin \theta \cos \theta +\sin^2 \theta +\cos^2 \theta +2\sin \theta \cos \theta}{\sin^2 \theta - \cos^2 \theta} \\\\\Rightarrow \frac{\sin^2 \theta +\cos^2 \theta +\sin^2 \theta +\cos^2 \theta}{\sin^2 \theta - \cos^2 \theta} \\\\\Rightarrow \frac{2\sin^2 \theta +2\cos^2 \theta}{\sin^2 \theta - \cos^2 \theta}$

$\Rightarrow \frac{2(\sin^2 \theta +\cos^2 \theta)}{\sin^2 \theta - (1-\sin^2 \theta)}\\\\\Rightarrow \frac{2(\sin^2 \theta +\cos^2 \theta)}{\sin^2 \theta - 1+\sin^2 \theta}\\\\\Rightarrow \frac{2(\sin^2 \theta +\cos^2 \theta)}{2\sin^2 \theta - 1}\\\\\Rightarrow \frac{2\times 1}{2\sin^2 \theta - 1}\\\\\Rightarrow \frac{2}{2\sin^2 \theta - 1}$

The given value is correct because L.H.S = R.H.S.

$\bold{\frac{\sin \theta -\cos \theta}{\sin \theta + \cos \theta} + \frac{\sin \theta +\cos \theta}{\sin \theta - \cos \theta} \ = \frac{2}{2\sin^2 \theta-1} }$