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Trigonometric Functions — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSTrigonometric Functions4 Marks
Question
Prove the following identities $:\frac{(1+\cot\text{x}+\tan\text{x})(\sin\text{x}+\cos\text{x})}{\sec^3\text{x}-\text{cosec}^3\text{ x}}=\sin^2\text{x}\cos^2\text{x}$
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Answer

$\text{L.H.S}=\frac{(1+\cot\text{x}+\tan\text{x})(\sin\text{x}+\cos\text{x})}{\sec^3\text{x}-\text{cosec}^3\text{ x}}$ $=\frac{\Big(1+\frac{\cos\text{x}}{\sin\text{x}}+\frac{\sin\text{x}}{\cos\text{x}}\Big)}{\Big(\frac{1}{\cos^3\text{x}}-\frac{1}{\sin^3\text{x}}\Big)}$
$\left(\begin{array}{c}\because\cot\text{x}=\frac{\cos\text{x}}{\sin\text{x}},\tan\text{x}=\frac{\sin\text{x}}{\cos\text{x}}\\ \sec\text{x}=\frac{1}{\cos\text{x}},\text{cosec }\text{x}=\frac{1}{\sin\text{x}}\end{array}\right)$
$=\Bigg(\frac{1+\frac{\cos^2\text{x}+\sin^2\text{x}}{\sin\text{x}\cos\text{x}}}{\frac{\sin^3\text{x}-\cos^3\text{x}}{\cos^3\text{x}\sin^3\text{x}}}\Bigg)(\sin\text{x}-\cos\text{x})$
$=\frac{(\sin\text{x}\cos\text{x}+1)\sin^3\text{x}\cos^3\text{x}}{\sin\text{x}\cos\text{x}.(\sin^3\text{x}-\cos^3\text{x})}\cdot(\sin\text{x}-\cos\text{x})$
$(\because\sin^2\text{x}+\cos^2\text{x}=1)$
$=\frac{(1+\sin\text{x}\cos\text{x})\sin^2\text{x}\cos^2\text{x}.(\sin\text{x}-\cos\text{x})}{(\sin\text{x}-\cos\text{x})(\sin^3\text{x}+\cos^2\text{x}+\sin\text{x}\cos\text{x})}$
$(\because\text{a}^3-\text{b}^3=(\text{a-b})(\text{a}^2+\text{b}^2+\text{ab})$
$=\frac{(1+\sin\text{x}\cos\text{x}).\sin^2\text{x}\cos^2\text{x}}{1+\sin\text{x}\cos\text{x}}$
$=\sin^2\text{x}\cos^2\text{x}$ $=\text{R.H.S}$
$\text{Proved}$

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