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Trigonometric Ratios Of Compounds — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSTrigonometric Ratios Of Compounds1 Mark
MCQ
If $\alpha$ and $\beta$ are acute angles satisfying $\cos2\alpha=\frac{3\cos2\beta-1}{3-\cos2\beta},$ then $\tan\alpha=$
  • ✓
    $\sqrt{2}\tan\beta$
  • B
    $\frac{1}{\sqrt{2}}\tan\beta$
  • C
    $\sqrt{2}\cot\beta$
  • D
    $\frac{1}{\sqrt{2}}\cot\beta$
✓

Answer

Correct option: A.
$\sqrt{2}\tan\beta$
Given:
$\cos2\alpha=\frac{3\cos2\beta-1}{3-\cos2\beta}$
$\Rightarrow\frac{\cos2\alpha-1}{\cos2\alpha+1}=\frac{(3\cos2\beta-1)-(3-\cos2\beta)}{(3\cos2\beta-1)+(3-\cos2\beta)}$ (Using componendo and dividendo)
$\Rightarrow\frac{\cos2\alpha-1}{\cos2\alpha+1}=\frac{4\cos2\beta-4}{2\cos2\beta+2}$
$\Rightarrow​-​=\frac{1-\cos^2\alpha}{1+\cos2\alpha}=\frac{-4(1-\cos2\beta)}{2(1+\cos2\beta)}$
$\Rightarrow\frac{1-\cos2\alpha}{1+\cos2\alpha}=\frac{2(1-\cos2\beta)}{(1+\cos2\beta)}$
$\Rightarrow\frac{2\sin^2\alpha}{2\cos^2\alpha}=\frac{2(2\sin^2\beta)}{2\cos^2\beta}$
$\Rightarrow\tan^2\alpha=2\tan^2\beta$
$\therefore\tan\alpha=\sqrt{2}\tan\beta$

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