Question

Prove that:
(i) $\sqrt{\frac{1-\sin A}{1+\sin A}}=\sec A-\tan A$
(ii) [tex](1+\tan^{2} A)+(1+\frac{1}{\tan^{2} A})=\frac{1}{\sin^{2}A-\sin^{4}A​

Answer 1

$\sf \Large Solution!!$

$\sf \large Part\;(i)$

$\sf L.H.S=\sqrt{\frac{1-\sin A}{1+\sin A}}\times \frac{\sqrt{1-\sin A}}{\sqrt{1-\sin A}}\quad [Multiplying\;and\;dividing\;by\;\sqrt{1-\sin A}]$

$\sf =\frac{1-\sin A}{\sqrt{1-\sin^{2}A}}=\frac{1-\sin A}{\cos A}\quad [\because 1-\sin^{2}A=\cos^{2}A]$

$\sf =\frac{1}{\cos A}-\frac{\sin A}{\cos A}=\sec A-\tan A$

$\sf L.H.S=R.H.S$

$\sf Hence,\;proved.$

$\sf \large Part\;(ii)$

$\sf L.H.S=(1+\tan^{2} A)+(1+\frac{1}{\tan^{2} A})$

$\sf =\sec^{2}A+(1+\cot^{2}A)$

$\sf =\sec^{2}A+\cosec^{2}A$

$\sf \frac{1}{\cos^{2}A}+\frac{1}{\sin^{2}A}=\frac{\sin^{2}A+\cos^{2}A}{\cos^{2}A\sin^{2}A}=\frac{1}{\cos^{2}A\sin^{2}A}$

$\sf =\frac{1}{(1-\sin^{2}A)\sin^{2}A}=\frac{1}{\sin^{2}A-\sin^{4}A}$

$\sf L.H.S=R.H.S$

$\sf Hence,\;proved.$