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Trigonometric Identities — Maths STD 10 — Question

Maharashtra BoardEnglish MediumSTD 10MathsTrigonometric Identities2 Marks
Question
Prove the following trigonometric identities.
$\frac{\sec\text{A}-\tan\text{A}}{\sec\text{A}+\tan\text{A}}=\frac{\cos^2\text{A}}{(1+\sin\text{A})^2}$
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Answer

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

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