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Inverse Trigonometric Functions — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsInverse Trigonometric Functions1 Mark
MCQ
If $\cos^{-1}\frac{\text{x}}{3}+\cos^{-1}\frac{\text{y}}{2}=\frac{\theta}{2},$ then, $4\text{x}^2-12\text{xy}\cos^2\frac{\theta}{2}+9\text{y}^2=$
  • A
    $36$
  • B
    $36-36\cos\theta$
  • ✓
    $18-18\cos\theta$
  • D
    $18+18\cos\theta$
✓

Answer

Correct option: C.
$18-18\cos\theta$
$\cos^{-1}\text{x}+\cos^{-1}\text{y}=\cos^{-1}\Big(\text{xy}\sqrt{1-\text{x}^2}\sqrt{1-\text{y}^2}\Big)$
$\Rightarrow\cos^{-1}\frac{\text{x}}{3}+\cos^{-1}\frac{\text{y}}{2}=\frac{\theta}{2}$
$\Rightarrow\cos^{-1}\Bigg(\frac{\text{x}}{3}\times\frac{\text{y}}{2}-\sqrt{1-\Big(\frac{\text{x}}{3}\Big)^2}\sqrt{1-\Big(\frac{\text{y}}{2}\Big)^2}\Bigg)=\frac{\theta}{2}$
$\Rightarrow\frac{\text{xy}}{6}-\sqrt{1-\Big(\frac{\text{x}^2}{9}\Big)}\sqrt{1-\Big(\frac{\text{y}^2}{4}\Big)}=\cos\frac{\theta}{2}$
$\Rightarrow\frac{\text{xy}-6\cos\frac{\theta}{2}}{6}=\frac{\sqrt{9-\text{x}^2}\sqrt{4-\text{y}^2}}{6}$
$\Rightarrow\text{xy}-6\cos\frac{\theta}{2}=\sqrt{9-\text{x}^2}\sqrt{4-\text{y}^2}$
Taking square on both sides,
$\Rightarrow\text{x}^2\text{y}^2-12\text{xy}\cos\frac{\theta}{2}+36\cos^2\frac{\theta}{2}=\big(9-\text{x}^2\big)\big(4-\text{y}^2\big)$
$\Rightarrow\text{x}^2\text{y}^2-12\text{xy}\cos\frac{\theta}{2}+36\cos^2\frac{\theta}{2}=36-9\text{y}^2-4\text{x}^2+\text{x}^2\text{y}^2$
$\Rightarrow4\text{x}^2+9\text{y}^2-12\text{xy}\cos^2\frac{\theta}{2}=36\Big(1-\cos^2\frac{\theta}{2}\Big)$
$\Rightarrow4\text{x}^2+9\text{y}^2-12\text{xy}\cos^2\frac{\theta}{2}=36\Big(1-\frac{1+\cos\theta}{2}\Big)$
$\Rightarrow4\text{x}^2+9\text{y}^2-12\cos^2\frac{\theta}{2}=18-18\cos\theta$

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