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Trigonometric Identities — Maths STD 10 — Question

CBSE BoardEnglish MediumSTD 10MathsTrigonometric Identities5 Marks
Question
Prove the following trigonometric identities.
$\frac{\cot^2\text{A}(\sec\text{A}-1)}{1+\sin\text{A}}=\sec^2\text{A}\Big(\frac{1-\sin\text{A}}{1+\sin\text{A}}\Big)$
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Answer

We have to prove $\frac{\cot^2\text{A}(\sec\text{A}-1)}{1+\sin\text{A}}=\sec^2\text{A}\Big(\frac{1-\sin\text{A}}{1+\sin\text{A}}\Big)$
We know that, $\sin^2\text{A}+\cos^2\text{A}=1$
So,
$\text{L.H.S}=\frac{\cot^2\text{A}(\sec\text{A}-1)}{1+\sin\text{A}}=\sec^2\text{A}\Big(\frac{1-\sin\text{A}}{1+\sin\text{A}}\Big)$
$=\frac{\frac{\cos^2\text{A}}{\sin^\text{2}\text{A}}\big(\frac{1}{\cos\text{A}}-1\Big)}{1+\sin\text{A}}$
$=\frac{\frac{\cos^2\text{A}}{\sin^2\text{A}}\frac{1-\cos\text{A}}{\cos\text{A}}}{1+\sin\text{A}}$
$=\frac{\cos\text{A}(1-\cos\text{A})}{\sin^2\text{A}(1+\sin\text{A})}$
$=\frac{\cos\text{A}(1-\cos\text{A})}{(1-\cos^2\text{A})(1+\sin\text{A})}$
$=\frac{\cos\text{A}(1-\cos\text{A})}{(1-\cos\text{A})(1+\cos\text{A})(1+\sin\text{A})}$
$=\frac{\cos\text{A}}{(1+\cos\text{A})(1+\sin\text{A})}$
$=\frac{\frac{1}{\sec\text{A}}}{\Big(1+\frac{1}{\sec\text{A}}\Big)(1+\sin\text{A})}$
$=\frac{\frac{1}{\sec\text{A}}}{\Big(\frac{\sec\text{A}+1}{\sec\text{A}}\Big)(1+\sin\text{A})}$
$=\frac{1}{(\sec\text{A}+1)(1+\sin\text{A})}$
Multiplying both the numerator ans denominator by $(1-\sin\text{A})$ we have
$=\frac{(1-\sin\text{A})}{(\sec\text{A}+1)(1+\sin\text{A})(1-\sin\text{A})}$
$=\frac{1-\sin\text{A}}{(\sec\text{A}+1)\cos^2\text{A}}$
$=\frac{(1-\sin\text{A})}{(\sec\text{A}+1)\cos^2\text{A}}$
$=\sec^2\text{A}\frac{(1-\sin\text{A})}{(\sec\text{A}+1)}$
$=\sec^2\text{A}\Big(\frac{1-\sin\text{A}}{1+\sec\text{A}}\Big)=\text{R.H.S}$
$\therefore\ \text{L.H.S}=\text{R.H.S}$

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