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Provide Solution For R.D.Sharma Maths Class 12 Chapter 3 Inverse Trigonometric Functions  Exercise Multiple Choice Questions Question 17 Maths Textbook Solution.

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Answer: 36 \sin ^{2} \theta

Hint: To find the variable value, separate the cos function.

Given: \cos ^{-1} \frac{x}{2}+\cos ^{-1} \frac{y}{3}=\theta

We have to compute 9 x^{2}-12 x y \cos \theta+4 y^{2}


We know that,

\cos ^{-1} x+\cos ^{-1} y=\cos ^{-1}\left(x y-\sqrt{1-x^{2}} \sqrt{1-y^{2}}\right)


\cos ^{-1}\left(\frac{x}{2} \frac{y}{3}-\sqrt{1-\frac{x^{2}}{4}} \sqrt{1-\frac{y^{2}}{9}}\right)=\theta

\left(\frac{x}{2} \frac{y}{3}-\sqrt{1-\frac{x^{2}}{4}} \sqrt{1-\frac{y^{2}}{9}}\right)=\cos \theta

\begin{aligned} &x y-\sqrt{4-x^{2}} \sqrt{9-y^{2}}=6 \cos \theta \\ &x y-6 \cos \theta=\sqrt{4-x^{2}} \sqrt{9-y^{2}} \end{aligned}

Squaring on both sides,

\begin{aligned} &x y-6 \cos \theta=\sqrt{4-x^{2}} \sqrt{9-y^{2}} \\ &x^{2} y^{2}+36 \cos ^{2} \theta-12 x y \cos \theta=\left(4-x^{2}\right)\left(9-y^{2}\right) \\ &x^{2} y^{2}+36 \cos ^{2} \theta-12 x y \cos \theta=36-4 y^{2}-9 x^{2}+x^{2} y^{2} \end{aligned}

\begin{aligned} &36-4 y^{2}-9 x^{2}=36 \cos ^{2} \theta-12 x y \cos \theta \\ &36-36 \cos ^{2} \theta=-12 x y \cos \theta+4 y^{2}+9 x^{2} \\ &36\left(1-\cos ^{2} \theta\right)=-12 x y \cos \theta+4 y^{2}+9 x^{2} \\ &9 x^{2}+4 y^{2}-12 x y \cos \theta=36 \sin ^{2} \theta \end{aligned}

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