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Trigonometric Ratios Of Compound Angles — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsTrigonometric Ratios Of Compound Angles3 Marks
Question
If $\tan\text{x}=\frac{\text{b}}{\text{a}},$ then find the value of $\sqrt{\frac{\text{a+b}}{\text{a}-\text{b}}}+\sqrt{\frac{\text{a}-\text{b}}{\text{a}+\text{b}}}$
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Answer

$\sqrt{\frac{\text{a+b}}{\text{a}-\text{b}}}+\sqrt{\frac{\text{a}-\text{b}}{\text{a}+\text{b}}}=\frac{(\text{a+b})+(\text{a}-\text{b})}{\sqrt{(\text{a}-\text{b})(\text{a+b})}}$
$=\frac{2\text{a}}{\sqrt{\text{a}^2-\text{b}^2}}$
$=\frac{2}{\sqrt{1-\Big(\frac{\text{b}}{\text{a}}\Big)^2}}$
$=\frac{2}{\sqrt{\tan^2\text{x}}}$ $\Big[\because\tan\text{x}=\frac{\text{b}}{\text{a}}\Big]$
$=\frac{2\cos\text{x}}{\sqrt{\cos^2\text{x}}\sin^2\text{x}}$
$=\frac{2\cos\text{x}}{\sqrt{\cos2\text{x}}}$

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