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Need solution for RD Sharma maths class 12 Chapter Inverse Trigonometric functions exercise 3.8 question 2 sub question (i)

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We convert in the form of \tan (\alpha + \beta) so we know the formula of it.


We have to prove that \tan \left(\cos ^{-1} \frac{4}{5}+\tan ^{-1} \frac{2}{3}\right)=\frac{17}{6} 


 LHS= \tan \left(\cos ^{-1} \frac{4}{5}+\tan ^{-1} \frac{2}{3}\right)

Let suppose that \cos ^{-1}\frac{4}{5}=\alpha and \tan ^{-1}\frac{2}{3}=\beta,

                \begin{aligned} &\cos \alpha=\frac{4}{5} \text { and } \tan \beta=\frac{2}{3} \\ &\tan (\alpha+\beta)=\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta} \end{aligned}

But we have not \tan \alpha. We have only \tan \beta. So, we convert \cos \alpha into \tan \alpha.

                \cos \alpha =\frac{4}{5}=\frac{B}{H}

Let, in \Delta ABC,  angle CAB = \alpha and right angle at B.


\begin{aligned} &\\ &(A C)^{2}=(A B)^{2}+(B C)^{2} \\ &(5)^{2}=4^{2}+(B C)^{2} \\ &(B C)^{2}=25-16 \\ \end{aligned}

BC =\pm 9

BC = 3

\tan \alpha = \frac{3}{4}


                                                                                                            [\because \tan \alpha = \frac{3}{4} \; and\: \tan \beta = \frac{2}{3} ]

\begin{gathered} \tan (\alpha+\beta)=\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta} \\ =\frac{\frac{3}{4}+\frac{2}{3}}{1-\frac{3}{4} \times \frac{2}{3}} \\ =\frac{\frac{9+8}{12-6}}{12}=\frac{17}{6}=\mathrm{RHS} \\ \tan \left(\cos ^{-1} \frac{4}{5}+\tan ^{-1} \frac{2}{3}\right)=\frac{17}{6} \end{gathered}                                           [ we take LCM of 3,4]

Hence proved.

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