Maharashtra BoardEnglish MediumSTD 12 ScienceMathsInverse Trigonometric Functions5 Marks
Question
If $\Big(\sin^{-1}\text{x}\Big)^2+\Big(\cos^{-1}\text{x}\Big)^2=\frac{175\pi^2}{36},$ find x.
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Answer
$\Big(\sin^{-1}\text{x}\Big)^2+\Big(\cos^{-1}\text{x}\Big)^2=\frac{175\pi^2}{36}$ $\Rightarrow\Big(\sin^{-1}\text{x}\Big)^2+\Big(\frac{\pi}{2}-\sin^{-1}\text{x}\Big)^2=\frac{175\pi^2}{36}$ Let $\sin^{-1}\text{x}=\text{y}$ $\therefore\ (\text{y})^2+\Big(\frac{\pi}{2}-\text{y}\Big)^2=\frac{17\pi^2}{36}$ $\Rightarrow\text{y}^2+\frac{\pi^2}{4}+\text{y}^2-2\times\frac{\pi}{2}\times\text{y}=\frac{17\pi^2}{36}$ $\Rightarrow2\text{y}^2-\pi\text{y}=\frac{2\pi^2}{9}$$\Rightarrow18\text{y}^2-9\pi\text{y}-2\pi^2=0$
$\Rightarrow18\text{y}^2-12\pi\text{y}+3\pi\text{y}-2\pi\text{x}^2=0$
$\Rightarrow6\text{y}(3\text{y}-2\pi)+\pi(3\text{y}+2\pi)=0$
$\Rightarrow(3\text{y}-2\pi)(6\text{y}+\pi)=0$
$\Rightarrow\text{y}=-\frac{\pi}{6}$
[Neglecting $\text{y}=\frac{2}{3}\pi$ as it is not satisfying the question]$\therefore\text{x}=\sin\text{y}=\sin\Big(-\frac{\pi}{6}\Big)=-\frac{1}{2}$
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