Prove the following trigonometric identities. (secA- )^2= 1- 1+ - Vidyadip

Question

Prove the following trigonometric identities.
$(\text{sec}\text{A}-\tan\text{A})^2=\frac{1-\sin\text{A}}{1+\sin\text{A}}$

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Answer

$\text{L.H.S}=(\sec\text{A}-\tan\text{A})^2$
$\Rightarrow\ \Big[\frac{1}{\cos\text{A}}-\frac{\sin\text{A}}{\cos\text{A}}\Big]^2\Rightarrow\ \frac{(1-\sin\text{A})^2}{\cos^2\text{A}}$
$\Rightarrow\ \frac{(1-\sin\text{A})^2}{1-\sin^2\text{A}} \big[\because\ 1-\sin^2\text{A}=\cos^2\text{A}\big]$
$\Rightarrow\ \frac{(1-\sin\text{A})^2}{(1-\sin\text{A})(1+\sin\text{A})} \big[\because \text{a}^2-\text{b}^2=(\text{a}-\text{b})(\text{a}+\text{b})\big]$
$=\frac{1-\sin\text{A}}{1+\sin\text{A}}$
$\therefore \text{L.H.S}=\text{R.H.S}$
Hence proved.

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