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Trigonometric Functions — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsTrigonometric Functions1 Mark
MCQ
If $\tan\frac{\text{x}}{2}=\sqrt{\frac{1-\text{e}}{1+\text{e}}}\tan\frac{\alpha}{2}$ then $\cos\alpha=$
  • A
    $1-\text{e}\cos(\cos\text{x}+\text{e})$
  • B
    $\frac{1+\text{e}\cos\text{x}}{\cos\text{x}-\text{e}}$
  • C
    $\frac{1-\text{e}\cos\text{x}}{\cos\text{x}-\text{e}}$
  • ✓
    $\frac{\cos\text{x}-\text{e}}{1-\text{e}\cos\text{x}}$
✓

Answer

Correct option: D.
$\frac{\cos\text{x}-\text{e}}{1-\text{e}\cos\text{x}}$
Given:
$\tan\frac{​​\text{x}}{2}=\sqrt{\frac{1-\text{e}}{1+\text{e}}}\tan\frac{\alpha}{2}$
$\Rightarrow\frac{\tan\frac{\text{x}}{2}}{\tan\frac{\alpha}{2}}=\sqrt{\frac{1-\text{e}}{1+\text{e}}}$
Squaring both sides, we get,
$\frac{\tan^2\frac{\text{x}}{2}}{\tan^2\frac{\alpha}{2}}=\frac{1-\text{e}}{1+\text{e}}$
$\Rightarrow\tan^2\frac{\alpha}{2}(1-\text{e})=\tan^2\frac{\text{x}}{2}(1+\text{e})$
$\Rightarrow\frac{\sin^2\frac{\alpha}{2}}{\cos^2\frac{\alpha}{2}}(1-\text{e})=\frac{\sin^2\frac{\text{x}}{2}}{\cos^2\frac{\text{x}}{2}}(1+\text{e})$
$\Rightarrow\frac{\frac{1}{2}(1-\cos\alpha)}{\frac{1}{2}(1+\cos\alpha)}(1-\text{e})=\frac{\frac{1}{2}(1-\cos\text{x})}{\frac{1}{2}(1+\cos\text{x})}(1+\text{e})$
$\Rightarrow(1-\cos\alpha)(1+\cos\text{x})(1-\text{e})=(1+\cos\alpha)(1-\cos\text{x})(1+\text{x})$
$\Rightarrow(1+\cos\text{x})(1-\text{e})-\cos\alpha(1+\cos\text{x})(1-\text{e})\\=(1-\cos\text{x})(1+\text{e})+\cos\alpha(1-\cos\text{x})(1+\text{e})$
$\Rightarrow\cos\alpha\big\{(1+\cos\text{x})(1-\text{e})+(1-\cos\text{x})(1+\text{e})\big\}\\=(1+\cos\text{x})(1-\text{e})-(1-\cos\text{x})(1+\text{e})$
$\Rightarrow\cos\alpha=\frac{2\cos\text{x}-2\text{e}}{2-2\text{e}\cos\text{x}}=\frac{\cos\text{x}-\text{e}}{1-\text{e}\cos\text{x}}$

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