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  • Need solution for RD Sharma Maths Class 12 Chapter 3 Inverse Trigonometric Functions Excercise Fill in the Blanks Question 6

Need solution for RD Sharma Maths Class 12 Chapter 3 Inverse Trigonometric Functions Excercise Fill in the Blanks Question 6

Answers (1)

best_answer

Answer:

                \frac{3\pi}{4}

Hint:

You must know the value of inverse trigonometric function.

Given:

\tan^{-1}2+\tan^{-1}3

Solution:

                \tan^{-1}2+\tan^{-1}3

\Rightarrow         =\left [ \frac{\pi}{2}-\cot^{-1} 2\right ]+\left [ \frac{\pi}{2}-\cot^{-1} 3\right ]

\Rightarrow         =\frac{\pi}{2}+\frac{\pi}{2}-\left [ \cot^{-1}2+\cot^{-1}3 \right ]

\Rightarrow         =\frac{2\pi}{2}-\left [ \tan^{-1}\left ( \frac{1}{2} \right ) +\tan^{-1}\left ( \frac{1}{3} \right )\right ]

\Rightarrow         =\pi-\left [ \tan^{-1}\left ( \frac{\frac{1}{2}+\frac{1}{3}}{1-\left ( \frac{1}{2} \right )\left ( \frac{1}{3} \right )} \right ) \right ]                                             \left [ \because \tan^{-1}a+\tan^{-1}b =\tan^{-1}\left ( \frac{a+b}{1-ab} \right )\right ]

\Rightarrow         =\pi-\left [ \tan^{-1}\left ( \frac{\frac{5}{6}}{\frac{5}{6}} \right ) \right ]

\Rightarrow         =\pi-\left [ \tan^{-1}\left ( 1\right ) \right ]                                                                \left [ 1=tan^{-1}(tan\frac{\pi}{4}) \right ]

\Rightarrow         =\pi-\frac{\pi}{4}

\Rightarrow         =\frac{4\pi-\pi}{4}

\Rightarrow         =\frac{3\pi}{4}

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