Question

Prove that :
$\sqrt{\frac{1-sin\theta }{1+sin\theta }}=sec\theta -tan\theta$

Answer 1

L.H.S. = $\sqrt{\frac{1-sin\theta }{1+sin\theta }}$

$\Rightarrow$ L.H.S. = $\sqrt{\frac{1-sin\theta }{1+sin\theta }\times \frac{1-sin\theta }{1-sin\theta }}$

$\Rightarrow$ L.H.S. = $\sqrt{\frac{(1-sin\theta {)}^2}{1-si{n}^2\theta }}=\sqrt{\frac{(1-sin\theta {)}^2}{co{s}^2\theta }}$

$\Rightarrow$ L.H.S. = $\sqrt{{\left(\frac{1-sin\theta }{cos\theta }\right)}^2}=\frac{1-sin\theta }{cos\theta }\frac{1}{cos\theta }\frac{sin\theta }{cos\theta }=sec\theta -tan\theta =R.H.S.$