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Trigonometry – II — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsTrigonometry – II2 Marks
MCQ
If $\tan x=\frac{b}{a}$, then $\sqrt{\frac{a+b}{a-b}}+\sqrt{\frac{a-b}{a+b}}=$
  • A
    $\frac{2 \sin x}{\sqrt{\sin 2 x}}$
  • ✓
    $\frac{2 \cos x}{\sqrt{\cos 2 x}}$
  • C
    $\frac{2 \cos x}{\sqrt{\sin 2 x}}$
  • D
    $\frac{2 \sin x}{\sqrt{\cos 2 x}}$
✓

Answer

Correct option: B.
$\frac{2 \cos x}{\sqrt{\cos 2 x}}$
(B)
Given that, $\tan x=\frac{ b }{ a }$
$\therefore \sqrt{\frac{a+b}{a-b}}+\sqrt{\frac{a-b}{a+b}}=\sqrt{\frac{1+\frac{b}{a}}{1-\frac{b}{a}}}+\sqrt{\frac{1-\frac{b}{a}}{1+\frac{b}{a}}}$
$=\frac{2}{\sqrt{1-\frac{b^2}{a^2}}}=\frac{2}{\sqrt{1-\tan ^2 x}}$
$=\frac{2}{\sqrt{1-\frac{\sin ^2 x}{\cos ^2 x}}}=\frac{2 \cos x}{\sqrt{\cos 2 x}}$

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