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Inverse Trigonometric Functions — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsInverse Trigonometric Functions3 Marks
Question
Write the value of $\sin\big(\cot^{-1}\text{x}\big).$
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Answer

We know$\cot^{-1}\text{x}=\tan^{-1}\frac{1}{\text{x}}$
Now, we have $\sin\big(\cot^{-1}\text{x}\big)=\sin\Big(\tan^{-1}\frac{1}{\text{x}}\Big)$ $=\sin\Bigg[\sin^{-1}\Bigg(\frac{\frac{1}{\text{x}}}{\sqrt{1+\frac{1}{\text{x}^2}}}\Bigg)\Bigg]$ $\Big[\because\ \tan^{-1}\text{x}=\sin^{-1}\Big(\frac{\text{x}}{\sqrt{1+\text{x}}}\Big)\Big]$ $=\sin\Bigg[\sin^{-1}\Bigg(\frac{\frac{1}{\text{x}}}{\frac{\sqrt{\text{x}^2+1}}{\text{x}}}\Bigg)\Bigg]$ $=\sin\bigg(\sin^{-1}\frac{1}{\sqrt{\text{x}^2+1}}\bigg)$ $=\frac{1}{\sqrt{\text{x}^2+1}}$ $\big[\because\ \sin\big(\sin^{-1}\text{x}=\text{x}\big)\big]$ Hence, $\sin\big(\cot^{-1}\text{x}\big)=\frac{1}{\sqrt{\text{x}^2-1}}.$

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