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Trigonometric Identities — Maths STD 10 — Question

Maharashtra BoardEnglish MediumSTD 10MathsTrigonometric Identities4 Marks
Question
Prove the following trigonometric identities.
$\frac{\sin\text{A}}{\sec\text{A}+\tan\text{A}-1}+\frac{\cos\text{A}}{\text{cosec A}+\cot\text{A}-1}=1$
✓

Answer

$\text{L.H.S}=\frac{\sin\text{A}}{\frac{1}{\cos\text{A}}+\frac{\sin\text{A}}{\cos\text{A}}-1}+\frac{\cos\text{A}}{\frac{1}{\sin\text{A}}+\frac{\cos\text{A}}{\sin\text{A}}-1}$
$=\frac{\sin\text{A}}{\frac{1+\sin\text{A}-\cos\text{A}}{\cos\text{A}}}+\frac{\cos\text{A}}{\frac{1+\cos\text{A}-\sin\text{A}}{\sin\text{A}}}$
$=\frac{\sin\text{A}\cos\text{A}}{1+\sin\text{A}-\cos\text{A}}+\frac{\sin\text{A}\cos\text{A}}{1+\cos\text{A}-\sin\text{A}}$
$=\sin\text{A}\cos\text{A}\Big[\frac{1}{1+\sin\text{A}-\cos\text{A}}+\frac{1}{1+\cos\text{A}-\sin\text{A}}\Big]$
$=\sin\text{A}\cos\text{A}\Big[\frac{1+\cos\text{A}-\sin\text{A}+1+\sin\text{A}-\cos\text{A}}{(1+\sin\text{A}-\cos\text{A})(1+\cos\text{A}-\sin\text{A})}\Big]$
$=\sin\text{A}\cos\text{A}\Big[\frac{2}{1+\cos\text{A}-\sin\text{A}+\sin\text{A}+\sin\text{A}\cos\text{A}-\sin^2\text{A}-\cos\text{A}-\cos^2\text{A}+\cos\text{A}\sin\text{A}}\Big]$
$=\sin\text{A}\cos\text{A}\Big[\frac{2}{1-\sin^2\text{A}-\cos^2\text{A}+2\sin\text{A}\cos\text{A}}\Big]$
$=\sin\text{A}\cos\text{A}\Big[\frac{2}{1-(\sin^2\text{A}+\cos^2\text{A})+2\sin\text{A}\cos\text{A}}\Big]$
$=\sin\text{A}\cos\text{A}\Big[\frac{2}{1-1+2\sin\text{A}\cos\text{A}}\Big](\because \sin^2\text{A}+\cos^2\text{A}=1)$
$=\sin\text{A}\times\cos\text{A}\times\frac{2}{2\sin\text{A}\cos\text{A}}$
$=1=\text{R.H.S}$
$\text{L.H.S}=\text{R.H.S}$

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