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Prove that :
\cos^{-1}\left ( \frac{12}{13} \right )+\sin^{-1}\left ( \frac{3}{5} \right )= \sin^{-1}\left ( \frac{56}{65} \right )

 

 

 

 
 
 
 
 

Answers (1)

L.H.S: \cos^{-1}\left ( \frac{12}{13} \right )+\sin^{-1}\left ( \frac{3}{5} \right )
          = \tan^{-1}\left ( \frac{5}{12} \right )+\tan^{-1}\left ( \frac{3}{4} \right )
\Rightarrow \tan^{-1}\left ( \frac{\frac{5}{12}+\frac{3}{4}}{1-\frac{5}{12}\times \frac{3}{4}} \right )\Rightarrow \tan^{-1}\left ( \frac{\frac{20+36}{\not{48}}}{\frac{48-15}{\not{48}}} \right )
\Rightarrow \tan^{-1}\left ( \frac{56}{33} \right )                                 Let  \tan^{-1}\left ( \frac{56}{33} \right )= \theta \Rightarrow \tan \theta = \frac{56}{33}
                  \therefore \sin \theta = \tan \theta \cos \theta = \tan \theta \times \frac{1}{\sec \theta }
Hence proved                                                           = \frac{56}{33}\times \frac{1}{\sqrt{1+\left ( \frac{56}{33} \right )^{2}}}
                                                                     \rightarrow \sin \theta = \frac{56}{33}\times \frac{33}{\sqrt{\left ( 33 \right )^{2}+56}}= \frac{56}{65}
                                                                        \therefore \theta = \sin^{-1}\frac{56}{65}           

LHS= RHS

Posted by

Ravindra Pindel

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