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Trigonometric Functions — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsTrigonometric Functions3 Marks
Question
Prove that:
$\frac{\sin(\pi+\text{x})\cos\big(\frac{\pi}{2}+\text{x}\big)\tan\big(\frac{3\pi}{2}-\text{x}\big)\cot(2\pi-\text{x})}{\sin(2\pi-\text{x})\cos(2\pi+\text{x})\text{cosec}(-\text{x})\sin\big(\frac{3\pi}{2}-\text{x}\big)}=1$
✓

Answer

$\text{L.H.S}=\frac{\sin(\pi+\text{x})\cos\big(\frac{\pi}{2}+\text{x}\big)\tan\big(\frac{3\pi}{2}-\text{x}\big)\cot(2\pi-\text{x})}{\sin(2\pi-\text{x})\cos(2\pi+\text{x})\text{cosec}(-\text{x})\sin\big(\frac{3\pi}{2}-\text{x}\big)}$
$=\frac{\sin\text{x}(-\sin\text{x})\cot(-\cot\text{x})}{-\sin\text{x}\cos\text{x}(-\text{cosec})(-\cos\text{x})}$ $\begin{pmatrix}\because\tan(270^\circ-\text{x})=\cot\text{x}\\\&\sin(270^\circ=\text{x})=-\cos\text{x}\end{pmatrix}$
$=\frac{-\sin\text{x}\times\sin\text{x}\times\cos\text{x}\times\cos\text{x}\times\sin\text{x}}{-\sin\text{x}\times\cos\text{x}\times\sin\text{x}\times\sin\text{x}\text{x}\cos\text{x}}$ $\begin{pmatrix}\because\cot\text{x}=\frac{\cos\text{x}}{\sin\text{x}}\\\&\text{ cosec}\text{x}=\frac{1}{\sin\text{x}}\end{pmatrix}$
$= 1$
$\text{= R.H.S}$
$\text{Proved}$

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