Prove that

    5. \cos^{-1}\frac{4}{5} + \cos^{-1}\frac{12}{13} = \cos^{-1}\frac{33}{65}

Answers (1)

Take \cos^{-1}\frac{4}{5} = x  and \cos^{-1}\frac{12}{13} = y    and \cos^{-1}\frac{33}{65} = z

then we have,

\cos x = \frac{4}{5}

\sin x = \sqrt {1- \left ( \frac {4}{5} \right )^2} = \frac {3}{5}

Then we can write it as:

\tan x = \frac{\frac{3}{5}}{\frac{4}{5}} = \frac{3}{4}    or    x= \tan^{-1} \frac{3}{4}

\therefore \cos ^{-1} \frac{4}{5} = \tan^{-1} \frac{3}{4}                                   ...............(1)

Now, \cos^{-1}\frac{12}{13} = y  

\cos y = \frac{12}{13} \Rightarrow \sin y =\frac{5}{13}

\therefore \tan y = \frac{5}{12} \Rightarrow y = \tan^{-1} \frac{5}{12}

So,  \cos^{-1}\frac{12}{13} = \tan^{-1} \frac{5}{12}                        ...................(2)

Also we have similarly;

 \cos^{-1}\frac{33}{65} = z

Then,

\cos^{-1}\frac{33}{65} = \tan^{-1} \frac{56}{33}             ...........................(3)

Now, we have

L.H.S

\cos^{-1}\frac{4}{5} + \cos^{-1}\frac{12}{13}    so, using (1) and (2) we get,

 =\tan^{-1}\frac{3}{4} + \tan^{-1}\frac{5}{12}

 =\tan^{-1}\left ( \frac{\frac{3}{4}+ \frac{5}{12}}{1-\left ( \frac{3}{4}\times \frac{5}{12} \right )} \right )             \because \left [ \tan^{-1}x + \tan^{-1}y = \tan^{-1} \frac{x+y}{1- xy} \right ]

=\tan^{-1}\left ( \frac{36+20}{48-15} \right )

 =\tan^{-1}\left ( \frac{56}{33} \right )    or we can write it as;

 =\cos^{-1}\frac{33}{65} 

=  R.H.S.

Hence proved.

 

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