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Trigonometric Ratios Of Compounds — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSTrigonometric Ratios Of Compounds3 Marks
Question
Prove that: $\cot^2\text{x}-\tan^2\text{x}=4\cot2\text{x}\ \text{cosec}\ 2\text{x}$
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Answer

$\text{LHS}=\cot^2\text{x}=\tan^2\text{x}$ $=\frac{\cos^2\text{x}}{\sin^2\text{x}}-\frac{\sin^2\text{x}}{\cos^2\text{x}}$ $=\frac{(\cos^2\text{x})^2-(\sin^2\text{x})^2}{\sin^2\text{x}\cos^2\text{x}}$ $=\frac{(\cos^2\text{x}+\sin^2\text{x})(\cos^2\text{x}-\sin^2\text{x})}{(\sin\text{x}\cos\text{x}^2)}$ $[\because\text{a}^2-\text{b}^2-=(\text{a+b})(\text{a-b})]$ $=\frac{\cos2\text{x}}{\frac{1}{4}(2\sin\text{x}\cos\text{x})^2}$ $[\because\cos2\text{x}=\cos^2\text{x}-\sin^2\text{x}]$ $=\frac{4\cos2\text{x}}{\sin^22\text{x}}$ $=\frac{4\cos2\text{x}}{\sin^2\text{x}}.\frac{1}{\sin2\text{x}}$ $\Big[\because\text{cosec}\ \theta=\frac{1}{\sin\theta}\Big]$ $=4\cot1\text{x}.\text{cosec}2\text{x}=\text{RHS}$

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