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Prove that

6. $\cos^{-1}\frac{12}{13} + \sin^{-1}\frac{3}{5} = \sin^{-1}\frac{56}{65}$

Answers (1)

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Converting all terms in tan form;

Let $\cos^{-1}\frac{12}{13} = x$ , $\sin^{-1}\frac{3}{5} = y$ and $\sin^{-1}\frac{56}{65} = z$ .

now, converting all the terms:

$\cos^{-1}\frac{12}{13} = x$ or $\cos x = \frac{12}{13}$

We can write it in tan form as:

$\cos x = \frac{12}{13} \Rightarrow$ $\sin x = \frac{5}{13}$ .

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

or $\cos^{-1}\frac{12}{13} = \tan^{-1} \frac{5}{12}$ ................(1)

$\sin^{-1}\frac{3}{5} = y$ or $\sin y = \frac{3}{5}$

We can write it in tan form as:

$\sin y = \frac{3}{5} \Rightarrow$ $\cos y = \frac{4}{5}$

$\therefore \tan y =\frac{3}{4} \Rightarrow y = \tan^{-1} \frac{3}{4}$

or $\sin^{-1}\frac{3}{5} = \tan^{-1} \frac{3}{4}$ ......................(2)

Similarly, for $\sin^{-1}\frac{56}{65} = z$ ;

we have $\sin^{-1}\frac{56}{65} = \tan^{-1} \frac{56}{33}$ .............(3)

Using (1) and (2) we have L.H.S

$\cos^{-1}\frac{12}{13} + \sin^{-1}\frac{3}{5}$

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

On applying $\tan^{-1}x + \tan^{-1}y = \tan^{-1} \frac{x+y}{1-xy}$

We have,

$=\tan^{-1} \frac{\frac{5}{12}+\frac{3}{4}}{1-(\frac{5}{12}.\frac{3}{4})}$

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

$=\tan^{-1} (\frac{56}{33})$

$=\sin^{-1} (\frac{56}{65})$ ...........[Using (3)]

=R.H.S.

Hence proved.

Posted by

Divya Prakash Singh

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