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Name: Please Solve RD Sharma Class 12 Chapter Inverse Trigonometric Function Exercise 3.13 Question 5 Maths Textbook Solution.
Uploaded: Sat, 2021-07-10 16:40
Description: Please Solve RD Sharma Class 12 Chapter Inverse Trigonometric Function Exercise 3.13 Question 5 Maths Textbook Solution.

Please Solve RD Sharma Class 12 Chapter Inverse Trigonometric Function Exercise 3.13 Question 5 Maths Textbook Solution.

Answers (1)

Given:
$\cos^{-1}\frac{12}{13}+\sin^{-1}\frac{3}{5}$
Hint:
First, we convert $\cos^{-1}$ to $\sin^{-1}$ and then use $\sin \left ( a+b \right )$ formula.
Solution:
Let          $a= \cos^{-1}\frac{12}{13}$ and $b= \sin^{-1}\frac{3}{5}$
Finding $\sin a , \cos a$ .
Let          $a= \cos^{-1}\frac{12}{13}$
         $\cos a= \left (\frac{12}{13} \right )$
We know that
               $\sin a= \sqrt{1-\cos ^{2}}a$
                          $= \sqrt{1-\left ( \frac{12}{13} \right )^{2}}$
                           $= \sqrt{\frac{25}{169}}$
                           $= \frac{5}{13}$
Again, finding $\sin b, \cos b$
Let          $b= \sin^{-1}\left ( \frac{3}{5} \right )$
         $\Rightarrow \sin b= \frac{3}{5}$
We know that,
                $\cos b= \sqrt{1-\sin ^{2}b}$
         $\Rightarrow \cos b= \sqrt{1-\left ( \frac{3}{5} \right )^{2}}$
                           $= \sqrt{\frac{16}{25}}= \frac{4}{5}$
                  $\cos b= \frac{4}{5}$
We know that,
                $\sin \left ( a+b \right )= \sin a\cos b+\cos a\sin b$
Putting, $\sin a= \frac{5}{13},\cos a= \frac{12}{13}$
                $\sin b= \frac{3}{5},\cos b= \frac{4}{5}$
We have,
                $\sin \left ( a+b \right )= \frac{5}{13}\times \frac{4}{5}+\frac{12}{13}\times \frac{3}{5}$
                                       $= \frac{20}{65}+\frac{36}{65}$
               $\sin \left ( a+b \right )= \frac{56}{65}$
                $a+b =\sin^{-1} \frac{56}{65}$
Putting the value of $a,b$
                $\cos^{-1}\frac{12}{13}+\sin^{-1}\frac{3}{5}= \sin^{-1}\frac{56}{65}$
                LHS=RHS
Hence proved.