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  • Please Solve R.D.Sharma Class 12 Chapter 3 Inverse Trigonometric Functions Exercise Multiple Choice Questions Question 21 Maths Textbbok Solution.

Please Solve R.D.Sharma Class 12 Chapter 3 Inverse Trigonometric Functions Exercise Multiple Choice Questions Question 21 Maths Textbbok Solution.

Answers (1)

Answer: \frac{-24}{25}

Hint: Separate cos function into simple form.

Given: \sin \left(2 \cos ^{-1}\left(\frac{-3}{5}\right)\right)

Solution:

\cos ^{-1}\left(\frac{-3}{5}\right)=\mathrm{x}, 0 \leq x \leq \pi

\cos x=\frac{-3}{5}

\sin x=\sqrt{1-\cos ^{2} x}

\sin x=\sqrt{1-\frac{9}{25}}

\sin x=\sqrt{\frac{16}{25}}

\sin x=\frac{4}{5}

Now,

\sin \left(2 \cos ^{-1}\left(\frac{-3}{5}\right)\right)=\sin 2 x

                                           \begin{aligned} &=2 \sin x \cos x \\ &=2\left(\frac{4}{5}\right)\left(\frac{-3}{5}\right) \\ &=\frac{-24}{25} \end{aligned}

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