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Please Solve R.D.Sharma Class 12 Chapter 3 Inverse Trigonometric Functions Exercise Multiple choice Questions Question 27 Maths Textbbok Solution.

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Answer: \frac{\pi}{4}

Hint: \tan ^{-1} x+\tan ^{-1} y=\tan ^{-1} \frac{x+y}{1-x y}

Given: \tan ^{-1}\left(\frac{a}{b+c}\right)+\tan ^{-1}\left(\frac{b}{c+a}\right)=?


\therefore \tan ^{-1}\left(\frac{a}{b+c}\right)+\tan ^{-1}\left(\frac{b}{c+a}\right)=\tan ^{-1}\left(\frac{\frac{a}{b+c}+\frac{b}{c+a}}{1-\left(\frac{a}{b+c}\right)\left(\frac{b}{c+a}\right)}\right)

                                                                      =\left(\tan ^{-1}\left(\frac{\frac{a c+a^{2}+b^{2}+b c}{(b+c)(c+a)}}{\frac{a c+c^{2}+b c}{(b+c)(c+a)}}\right)\right.

                                                                       =\tan ^{-1}\left(\frac{a c+c^{2}+b c}{a c+c^{2}+b c}\right)

                                                                     =\tan ^{-1}(1)  

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


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