MHT CET · Maths · Inverse Trigonometric Functions
If \(\cos ^{-1} x-\cos ^{-1} \frac{y}{3}=\alpha\), where \(-1 \leq x \leq 1\), \(-3 \leq y \leq 3, x \leq \frac{y}{3}\), then for all \(x, y\), \(9 x^2-6 x y \cos \alpha+y^2\) is equal to
- A \(\sin ^2 \alpha\)
- B \(3\sin ^2 \alpha\)
- C \(9\sin ^2 \alpha\)
- D \(\frac {4}{9}\sin ^2 \alpha\)
Answer & Solution
Correct Answer
(C) \(9\sin ^2 \alpha\)
Step-by-step Solution
Detailed explanation
If \(\cos ^{-1} \frac{x}{a}-\cos ^{-1} \frac{y}{b}=\theta\), then \(\frac{x^2}{\mathrm{a}^2}-\frac{2 x y}{\mathrm{ab}} \cos \theta+\frac{y^2}{\mathrm{~b}^2}=\sin ^2 \theta\) Given, \(\cos ^{-1} x-\cos ^{-1} \frac{y}{3}=\alpha\) Here, \(\mathrm{a}=1, \mathrm{~b}=3\)
\(\begin{aligned}
\therefore \quad & \frac{x^2}{1^2}-\frac{2 x y}{(1)(3)} \cos \alpha+\frac{y^2}{3^2}=\sin ^2 \alpha \\
& \Rightarrow x^2-\frac{2 x y}{3} \cos \alpha+\frac{y^2}{9}=\sin ^2 \alpha \\
& \Rightarrow 9 x^2-6 x y \cos \alpha+y^2=9 \sin ^2 \alpha
\end{aligned}\)
\(\begin{aligned}
\therefore \quad & \frac{x^2}{1^2}-\frac{2 x y}{(1)(3)} \cos \alpha+\frac{y^2}{3^2}=\sin ^2 \alpha \\
& \Rightarrow x^2-\frac{2 x y}{3} \cos \alpha+\frac{y^2}{9}=\sin ^2 \alpha \\
& \Rightarrow 9 x^2-6 x y \cos \alpha+y^2=9 \sin ^2 \alpha
\end{aligned}\)
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