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Prove that

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

Answers (1)


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)]


Hence proved.






Posted by

Divya Prakash Singh

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