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Trigonometric Functions — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsTrigonometric Functions5 Marks
Question
If $\tan\text{x}=\frac{\text{b}}{\text{a}},$ then find the value of $\sqrt{\frac{\text{a}+\text{b}}{\text{a}-\text{b}}}+\sqrt{\frac{\text{a}-\text{b}}{\text{a}+\text{b}}}$
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Answer

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

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