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Trigonometric Functions — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsTrigonometric Functions5 Marks
Question
Prove the following identities:
$\frac{1−\sin\text{x}\cos\text{x}}{\cos\text{x}(\sec\text{x}−\text{cosec}\text{x})}\cdot\frac{\sin^2\text{x}−\cos^2\text{x}}{\sin^3\text{x}\cos^3\text{x}}=\sin\text{x}$
✓

Answer

$\text{L.H.S}=\frac{1-\sin\text{x}\cos\text{x}}{\cos\text{x}\big(\sec\text{x}-\text{cosec}\text{x}\big)}.\frac{\sin^{2}\text{x}-\cos^{2}\text{x}}{\sin^{3}\text{x}+\cos^{3}\text{x}}$
$=\frac{1-\sin\text{x}\cos\text{x}}{\cos\text{x}\big(\frac{1}{\cos\text{x}}-\frac{1}{\sin\text{x}}\big)}\cdot\frac{\big(\sin\text{x}+\cos\text{x}\big)\big(\sin\text{x}-\cos\text{x}\big)}{\big(\sin\text{x}+\cos\text{x}\big)\big(\sin^{2}\text{x}+\cos^{2}\text{x}-\sin\text{x}\cos{\text{x}}\big)}$$$ $\begin{bmatrix}\text{Using a}^2-\text{b}^2=(\text{a - b)(a + b)} \\\text{and a}^3+\text{b}^3\text{(a}^2+\text{b}^2-\text{ab}) \end{bmatrix}$
$=\frac{\big(1-\sin\text{x}\cos\text{x}\big)}{\cos\text{x}\big(\frac{\sin\text{x}-\cos\text{x}}{\cos\text{x}\sin\text{x}}\big)}.\frac{\sin\text{x}-\cos\text{x}}{1-\sin\text{x}\cos\text{x}}$ $ \big(\because\sin^{2}\text{x}+\cos^{2}\text{x}=1\big)$
$=\frac{\cos\text{x}\sin\text{x}}{\cos\text{x}}$
$=\sin\text{x}$
$=\text{R.H.S}$
$\text{Proved}$

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