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Trigonometric Identities — Maths STD 10 — Question

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

$\text{LHS}=(1+\tan\theta+\cot\theta)(\sin\theta-\cos\theta)$
$=\Big(1+\frac{\sin\theta}{\cos\theta}+\frac{\cos\theta}{\sin\theta}\Big)(\sin\theta-\cos\theta)$
$=\Big(\frac{\cos\theta\sin\theta+\sin^2\theta+\cos^2\theta}{\cos\theta\sin\theta}\Big)(\sin\theta-\cos\theta)$
$=\frac{(\cos\theta\sin\theta+1)}{\cos\theta\sin\theta}(\sin\theta-\cos\theta)$
$\text{RHS}=\Big(\frac{\sec\theta}{\text{cosec}^2\theta}-\frac{\text{cosec }\theta}{\sec^2\theta}\Big)$
$=\Bigg(\frac{\frac{1}{\cos\theta}-\frac{1}{\sin\theta}}{\frac{1}{\sin^2\theta}\frac{1}{\cos^2\theta}}\Bigg)$
$=\Big(\frac{\sin^2\theta}{\cos\theta}-\frac{\cos^2\theta}{\sin\theta}\Big)=\frac{\sin^3\theta-\cos^3\theta}{\cos\theta\sin\theta}$
$=\frac{(\sin\theta-\cos\theta)\big(\sin^2\theta+\cos^2\theta+\cos\theta\sin\theta\big)}{\cos\theta\sin\theta}$ $\Big[\because\text{a}^3-\text{b}^3=(\text{a}-\text{b})\big(\text{a}^2+\text{ab}+\text{b}^2\big)\Big]$
$=\frac{(\sin\theta-\cos\theta)(1+\cos\theta\sin\theta)}{\cos\theta\sin\theta}$
$\therefore\ \text{R.H.S.}=\text{L.H.S.}$

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