Question

prove that:
1+cos theta/1-costheta=tan^2 theta/(sec theta-1)^2

Answer 1

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Answer 2

Answer:

To prove :

$\frac{1+cos\theta}{1- cos \theta}=\frac{tan^2 \theta}{(sec\theta - 1)^2}$

L.H.S :

$\frac{1+cos\theta}{1- cos \theta}$

$=\frac{1+cos\theta}{1- cos \theta}\times \frac{1-cos\theta}{1- cos \theta}$

$=\frac{1-cos^2\theta}{(1- cos \theta)^2}$

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

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

Since, sec² x - 1 = tan² x

$=\frac{tan^2\theta}{(sec\theta - 1)^2}$

= R.H.S.

Hence, proved.