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Trigonometric Functions — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSTrigonometric Functions3 Marks
Question
Prove that $:\frac{\text{cosec}(90^{\circ}+\text{x})+\cot(450^\circ+\text{x})}{\text{cosec}(90^\circ-\text{x})+\tan(180^\circ-\text{x})} +\frac{\tan\text{x}(180^\circ+\text{x})+\sec(180^\circ-\text{x})}{\tan(360^\circ+\text{x})-\sec(-\text{x})}=2$
✓

Answer

$\text{L.H.S}=\frac{\text{cosec}(90^{\circ}+\text{x})+\cot(450^\circ+\text{x})}{\text{cosec}(90^\circ-\text{x})+\tan(180^\circ-\text{x})}+\frac{\tan\text{x}(180^\circ+\text{x})+\sec(180^\circ-\text{x})}{\tan(360^\circ+\text{x})-\sec(-\text{x})}$
$=\frac{\sec\text{x}+\cot\Big(2\pi+\frac{\pi}{2}+\text{x}\Big)}{\sec\text{x}-\tan\text{x}}+\frac{\tan\text{x}-\sec\text{x}}{\tan\text{x}-\sec\text{x}}$
$\left(\begin{array}{c}\because\text{cosec}(90^\circ+\text{x})=\sec\text{x},\text{cosec}(90^\circ+\text{x})=\sec\text{x},\tan(180^\circ-\text{x})\\=-\tan\text{x}\sec(-\text{x})=\sec\text{x}\end{array}\right)$
$=\frac{\sec\text{x}+\cot\Big(\frac{\pi}{2}+\text{x}\Big)}{\sec\text{x}-\tan\text{x}}+1$
$(\because\cot(2\pi+\text{x})=\cot\text{x})$ $=\frac{\sec\text{x}-\tan\text{x}}{\sec\text{x}-\tan\text{x}}+1$
$\Big(\because\cot\Big(\frac{\pi}{2}+\text{x}\Big)=-\tan\text{x}\Big)$
$= 1 + 1$
$= 2$
$= \text{R.H.S}$
$\text{Proved}$

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