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Need solution for RD Sharma Maths Class 12 Chapter 3 Inverse Trigonometric Functions Excercise Fill in the Blanks Question 13

Answers (1)

Answer:

                \frac{\pi}{6}

Hint:

You must know the rules of inverse trigonometric function.

Given:

\tan^{-1}x+\tan^{-1}y=\frac{5\pi}{6}, then find cot^{-1}x+\cot^{-1}y

Solution:

Using inverse trigonometric rule,

We know

                \tan^{-1}x+\cot^{-1}x=\frac{\pi}{2}  and

                \tan^{-1}y+\cot^{-1}y=\frac{\pi}{2}

Adding both equations,

\begin{aligned} &\Rightarrow \quad \tan ^{-1} x+\cot ^{-1} x+\tan ^{-1} y+\cot ^{-1} y=\frac{\pi}{2}+\frac{\pi}{2} \\\\ &\Rightarrow \quad\left(\tan ^{-1} x+\tan ^{-1} y\right)+\left(\cot ^{-1} x+\cot ^{-1} y\right)=\frac{2 \pi}{2} \\ \\&\Rightarrow \quad \frac{5 \pi}{6}+\left(\cot ^{-1} x+\cot ^{-1} y\right)=\pi \\\\ &\Rightarrow \quad \cot ^{-1} x+\cot ^{-1} y=\pi-\frac{5 \pi}{6} \\ \\&\Rightarrow \quad \cot ^{-1} x+\cot ^{-1} y=\frac{6 \pi-5 \pi}{6} \end{aligned}        

\Rightarrow \cot^{-1}x+\cot^{-1}y=\frac{\pi}{6}        

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