Question

sin theta+cos theta÷sin theta-cos theta+sin theta-cos theta÷sin theta+cos theta=2÷1-cos²theta​

Answer 1

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Answer 2

Solution:

To prove:

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

Proof:

Taking L.H.S.,

= $\tt{\dfrac{sin\ \theta +cos\ \theta}{sin\ \theta -cos\ \theta} + \dfrac{sin\ \theta -cos\ \theta }{sin\ \theta +cos\ \theta } }$

On taking LCM, we get

= $\tt{\dfrac{(sin\ \theta +cos\ \theta)^{2}+(sin\ \theta -cos\ \theta)^{2} }{(sin\ \theta -cos\ \theta)(sin\ \theta+cos\ \theta)} }$

= $\tt{\dfrac{sin^{2} \ \theta +cos^{2} \ \theta + 2sin\ \theta cos\ \theta + sin^{2} \ \theta +cos^{2} \ \theta - 2sin\ \theta cos\ \theta }{sin^{2} \ \theta -cos^{2}\ \theta} }$

= $\tt{\dfrac{2sin^{2}\ \theta +2cos^{2}\ \theta}{sin^{2} \ \theta -cos^{2} \ \theta} }$

= $\tt{\dfrac{2(sin^{2} \ \theta +cos^{2} \ \theta)}{1-cos^{2}\ \theta -cos^{2}\ \theta} }$

= $\tt{\dfrac{2*1}{1-2cos^{2}\ \theta} }$

= $\tt{\dfrac{2}{1-2cos^{2}\ \theta} }$ = R.H.S.

Hence Proved

Identities used :-

sin²θ + cos²θ = 1

⇒ sin²θ = 1 - cos²θ