Question

Prove : $\frac{1}{sec\theta -tan\theta }+\frac{1}{sec\theta +tan\theta }=2sec\theta$

Answer 1

L.H.S.

$=\frac{1}{sec\theta -tan\theta }+\frac{1}{sec\theta +tan\theta }$

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

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

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

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

$=\frac{2cos\theta }{co{s}^2\theta }$

$=\frac{2}{cos\theta }$

$=2sec\theta$

$=$R.H.S.