MHT CET · Maths · Inverse Trigonometric Functions
If \(\sin ^{-1} x+\sin ^{-1} y=\frac{\pi}{3}\) and \(\cot ^{-1}\left(\frac{1}{x}\right)-\cot ^{-1}\left(\frac{1}{y}\right)=0\) then \(2 x^2+y^2-x y=\) \(\qquad\)
- A \(\frac{1}{4}\)
- B \(1\)
- C \(\frac{1}{2}\)
- D \(0\)
Answer & Solution
Correct Answer
(C) \(\frac{1}{2}\)
Step-by-step Solution
Detailed explanation
\(\cot ^{-1}\left(\frac{1}{x}\right)-\cot ^{-1}\left(\frac{1}{y}\right)=0 \implies \frac{1}{x}=\frac{1}{y} \implies x=y\) \(\sin ^{-1} x+\sin ^{-1} x=\frac{\pi}{3} \implies 2 \sin ^{-1} x=\frac{\pi}{3} \implies \sin ^{-1} x=\frac{\pi}{6} \implies x=\frac{1}{2}\)
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