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Trigonometric Functions — MATHS STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 ScienceMATHSTrigonometric Functions5 Marks
Question
Prove that $\Bigg|\sqrt{\frac{1-\sin\text{x}}{1+\sin\text{x}}}+\sqrt{\frac{1+\sin\text{x}}{1-\sin\text{x}}}\Bigg|$ $=-\frac{2}{\cos\text{x}},$ where $\frac{\pi}{2}<\text{x}<\pi$
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Answer

$\text{L.H.S}=\Bigg|\sqrt{\frac{1-\sin\text{x}}{1+\sin\text{x}}}+\sqrt{\frac{1+\sin\text{x}}{1-\sin\text{x}}}\Bigg|$
$=\Bigg|\frac{(\sqrt{1-\sin\text{x}})^2+(1+\sin\text{x})^2}{\sqrt{(1+\sin\text{x})(1-\sin\text{x})}}\Bigg|$
$=\Bigg|\frac{1-\sin\text{x}+1+\sin\text{x}}{\sqrt{1-\sin^2\text{x}}}\Bigg|$
$=\Big|\frac{2}{\cos\text{x}}\Big|$ $\Big(\because1-\sin^2\text{x}=\cos^2\text{x}\Rightarrow\sqrt{1-\sin^2\text{x}=\cos\text{x}}\Big)$
$=\frac{-2}{\cos\text{x}}$ $\Big(\frac{\pi}{2}<\text{x}<\pi\Rightarrow\cos\text{x}<0\Big)$
$=\text{R.H.S}$

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