If cot-1\(\left(\frac{-1}{5}\right)=x\) then sin x = ?A. \(\frac{1}{

1.	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
Answer» Correct Answer is (B) \(\frac{5}{\sqrt{26}}\) Given : \(cos^{-1}\frac{-1}{5}=x\) Since, x= \(cos^{-1}\frac{-1}{5}\) ⇒ \(cot x= \frac{-1}{5}=\frac{\text{adjacent side}}{\text{opposite side}}\) By pythagorus theroem , (Hypotenuse )² = (opposite side )²+ (adjacent side )² Therefore, Hypotenuse =\(\sqrt{26}\) ^⇒sin x =\(\frac{\text{opposite side}}{\text{hypotenuse side}}\) =\(\frac{5}{\sqrt{26}}\)

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

Answer»

Correct Answer is (B) \(\frac{5}{\sqrt{26}}\)

Given : \(cos^{-1}\frac{-1}{5}=x\)

Since, x= \(cos^{-1}\frac{-1}{5}\)

⇒ \(cot x= \frac{-1}{5}=\frac{\text{adjacent side}}{\text{opposite side}}\)

By pythagorus theroem ,

(Hypotenuse )² = (opposite side )²+ (adjacent side )²

Therefore, Hypotenuse =\(\sqrt{26}\)

^⇒sin x =\(\frac{\text{opposite side}}{\text{hypotenuse side}}\) =\(\frac{5}{\sqrt{26}}\)