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Trigonometric Identities — Maths STD 10 — Question

Maharashtra BoardEnglish MediumSTD 10MathsTrigonometric Identities4 Marks
Question
Prove the following trigonometric identities.
$(1+\cot\text{A}+\tan\text{A})(\sin\text{A}-\cos\text{A})=\frac{\sec\text{A}}{\text{cosec}^2\text{A}}-\frac{\text{cosec A}}{\sec^2\text{A}}=\sin\text{A}\tan\text{A}-\cot\text{A}\cos\text{A}$
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Answer

$\text{L.H.S}=(1+\cot\text{A}+\tan\text{A})(\sin\text{A}-\cos\text{A})$
$=\sin\text{A}-\cos\text{A}+\cot\text{A}\sin\text{A}-\cot\text{A}\cos\text{A}\\\ +\sin\text{A}\tan\text{A}-\tan\text{A}\cos\text{A}$
$$$=\sin\text{A}-\cos\text{A}+\frac{\cos\text{A}}{\sin\text{A}}\times\sin\text{A}-\cot\text{A}\cos\text{A}\\\ +\sin\text{A}\tan\text{A}-\frac{\sin\text{A}}{\cos\text{A}}\times\cos\text{A}$
$=\sin\text{A}-\cos\text{A}+\cos\text{A}-\cot\text{A}\cos\text{A}+\sin\text{A}\tan\text{A}-\sin\text{A}$
$=\sin\text{A}\cos\text{A}\cos\text{A}\cot\text{A}$
Solving:
$\frac{\sec\text{A}}{\text{cosec}^2\text{A}}-\frac{\text{cosec A}}{\sec^2\text{A}}$
$\frac{\frac{1}{\cos\text{A}}}{\frac{1}{\sin^2\text{A}}}-\frac{\frac{1}{\sin\text{A}}}{\frac{1}{\cos^2\text{A}}}$
$\frac{\sin^2\text{A}}{\cos\text{A}}-\frac{\cos^2\text{A}}{\sin\text{A}}$
$\frac{\sin^3\text{A}-\cos^3\text{A}}{\sin\text{A}\cos\text{A}}$
$=\sin\text{A}\times\frac{\sin\text{A}}{\cos\text{A}}-\cos\text{A}\times\frac{\cos\text{A}}{\sin\text{A}}$
$=\sin\text{A}\tan\text{A}-\cos\text{A}\cot\text{A}=\text{R.H.S}$
$\text{L.H.S}=\text{R.H.S}$

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