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  • Please solve RD Sharma Chapter Inverse trigonometric functions exercise 3.8 question 2 sub question (iii) maths text book solution

Please solve RD Sharma Chapter Inverse trigonometric functions exercise 3.8 question 2 sub question (iii) maths text book solution

Answers (1)

Hint:

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

Given:

We have to prove \tan \left(\sin ^{-1} \frac{5}{13}+\cos ^{-1} \frac{3}{5}\right)=\frac{63}{16} 

Solution:

       LHS = \tan \left(\sin ^{-1} \frac{5}{13}+\cos ^{-1} \frac{3}{5}\right)

\sin ^{-1} \frac{5}{13}=\alpha \; and \; \cos ^{-1} \frac{3}{5}=\beta

                \sin \alpha =\frac{5}{13} \; and \; \cos \beta = \frac{3}{5}

\tan (\alpha+\beta)=\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta}                                                                                                                         …(i)

But we have not \tan \alpha and  \tan \beta. So, we find out first then we put in equation (i)

                \sin \alpha = \frac{5}{13}=\frac{P}H{}

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

\begin{aligned} &\\ &(A C)^{2}=(A B)^{2}+(B C)^{2} \\ &(13)^{2}=AB^{2}+5^{2} \\ &(AB)^{2}=144 \\ \end{aligned}

AB =\pm 12                 [we will ignore the -ve sign because AB is a length and it can’t be -ve]     

AB =12

\therefore \tan \alpha = \frac{5}{12}

 

Let, in \Delta PQR,  angle RPQ = \beta and right angle at B.

\cot \beta =\frac{3}{5}=\frac{B}{P}

\begin{aligned} &\\ &(PR)^{2}=(QR)^{2}+(PQ)^{2} \\ &(5)^{2}=3^{2}+(QR)^{2} \\ &(QR)^{2}=16\\ \end{aligned}

QR = \pm 4

QR = 4

\therefore \tan \beta = \frac{4}{3}

Now,     \begin{aligned} \tan (\alpha+\beta)=& \frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta} \\ &=\frac{\frac{5}{12}+\frac{4}{3}}{1-\frac{5}{12} \times \frac{4}{3}} \end{aligned}                                                    [ put, \tan \alpha = \frac{5}{12} and \tan \beta = \frac{4}{3}]

                                                                                              [ we take LCM of 3 and 12]

                                        \begin{aligned} &=\frac{\frac{5+16}{12}}{\frac{36-20}{36}} \\ &=\frac{\frac{21}{12}}{\frac{16}{36}} \\ &=\frac{21}{12} \times \frac{36}{16} \\ &=\frac{21}{1} \times \frac{3}{16} \\ \tan (\alpha+\beta) &=\frac{63}{16}=\mathrm{RHS} \end{aligned}

Hence proved.

Posted by

infoexpert24

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