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Prove that

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

Answers (1)

best_answer

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.

Posted by

Divya Prakash Singh

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