JEE Mains · Maths · STD 12 - 2. inverse trigonometric function
If \(\frac{\sin ^{-1} x}{a}=\frac{\cos ^{-1} x}{b}=\frac{\tan ^{-1} y}{c} ; 0< x< 1,\) then the value of \(\cos \left(\frac{\pi c }{ a + b }\right)\) is
- A \(\frac{1-y^{2}}{y \sqrt{y}}\)
- B \(1-y^{2}\)
- C \(\frac{1-y^{2}}{1+y^{2}}\)
- D \(\frac{1-y^{2}}{2 y}\)
Answer & Solution
Correct Answer
(C) \(\frac{1-y^{2}}{1+y^{2}}\)
Step-by-step Solution
Detailed explanation
\(\frac{\sin ^{-1} x }{ r }= a , \frac{\cos ^{-1} x }{ r }= b , \frac{\tan ^{-1} y }{ r }= c\) So, \(a+b=\frac{\pi}{2 r}\) \(\cos \left(\frac{\pi c }{ a + b }\right)=\cos \left(\frac{\pi \tan ^{-1} y }{\frac{\pi}{2 r } r }\right)\) \(=\cos \left(2 \tan ^{-1} y \right),\) let…
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