Mark the tick against the correct answer in the following: If $\cot ^{-1}\left(\frac{-1}{5}\right)=x$ then $\sin x=?$ A. $\frac{1}{\sqrt{26}}$ B. $\frac{5}{\sqrt{26}}$ C. $\frac{1}{\sqrt{24}}$ D. none of these

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Mark the tick against the correct answer in the following: If $\cot ^{-1}\left(\frac{-1}{5}\right)=x$ then $\sin x=?$
A. $\frac{1}{\sqrt{26}}$
B. $\frac{5}{\sqrt{26}}$
C. $\frac{1}{\sqrt{24}}$
D. none of these

CBSE | Class 12 | Inverse Trigonometric Functions | Maths | RS Aggarwal

Mark the tick against the correct answer in the following: If $\cot ^{-1}\left(\frac{-1}{5}\right)=x$ then $\sin x=?$
A. $\frac{1}{\sqrt{26}}$
B. $\frac{5}{\sqrt{26}}$
C. $\frac{1}{\sqrt{24}}$
D. none of these

Solution:

Option(B) is correct.
Given: $\cot ^{-1} \frac{-1}{5}=x$
To Find: The value of $\sin \mathrm{x}$
Since, $x=\cot ^{-1} \frac{-1}{5}$
$\Rightarrow \cot \mathrm{x}=\frac{-1}{5}=\frac{\text { adjacent side }}{\text { opposite side }}$
By pythagorus theroem,
$(\text { Hypotenuse })^{2}=(\text { opposite side })^{2}+(\text { adjacent side })^{2}$
Therefore, Hypotenuse $=\sqrt{26}$
$\Rightarrow \sin x=\frac{\text { opposite side }}{\text { hypotenuse }}=\frac{5}{\sqrt{26}}$