Question

prove that cos^-1 (12/13) + sin^-1 (3/5) = sin^-1 (56/65)

Answer 1

We need to convert the terms on the LHS in $\tan^{-1}$ form.

$Let \, \, \cos^{-1} \left(\frac{12}{13} \right) = A \\ \\ \\ \implies \cos A = \frac{12}{13} \\ \\ \implies \sec A = \frac{1}{\cos A} = \frac{13}{12} \\ \\ \\ Now, \, \, \tan^2 A = \sec^2A-1 \\ \\ \\ \implies \tan^2A = \left( \frac{13}{12} \right)^2 - 1 \\ \\ \\ \implies \tan^2A = \frac{169}{144}-1 \\ \\ \\ \implies \tan^2 A = \frac{169-144}{144} \\ \\ \\ \implies \tan^2A = \frac{25}{144} \\ \\ \\ \implies \tan A = \frac{5}{12} \\ \\ \\ \implies A = \tan^{-1} \left( \frac{5}{12} \right)$

And, for the second term:


$Let \, \, \sin^{-1} \left(\frac{3}{5} \right) = B \\ \\ \\ \implies \sin B = \frac{3}{5} \\ \\ \implies cosec \, B = \frac{1}{\sin B} = \frac{5}{3} \\ \\ \\ Now, \, \, \cot^2 B = cosec^2 \, B-1 \\ \\ \\ \implies \cot^2B = \left( \frac{5}{3} \right)^2 - 1 \\ \\ \\ \implies \cot^2B = \frac{25}{9}-1 \\ \\ \\ \implies \cot^2 B = \frac{25-9}{9} \\ \\ \\ \implies \cot^2B = \frac{16}{9} \\ \\ \\ \implies \cot B = \frac{4}{3} \\ \\ \\ \implies \tan B = \frac{1}{\cot B} = \frac{3}{4}$


$\implies B = \tan^{-1} \left( \frac{3}{4} \right)$


Now, we will use the identity:


$\boxed{\tan^{-1}x+\tan^{-1}y = \tan^{-1} \left( \frac{x+y}{1-xy} \right) \, \, \, for \, \, xy\ \textless \ 1}$

Let us consider our LHS:

$LHS \\ \\ \\ = \cos^{-1} \left( \frac{12}{13} \right) + \sin^{-1} \left( \frac{3}{5} \right) \\ \\ \\ = \tan^{-1} \left(\frac{5}{12} \right) + \tan^{-1} \left( \frac{3}{4} \right) \\ \\ \\ Here \, \, \frac{5}{12} \times \frac{3}{4} = \frac{5}{16} \ \textless \ 1 \\ \\ \\ = \tan^{-1} \left( \frac{\frac{5}{12}+\frac{3}{4}}{1-\frac{5}{12} \times \frac{3}{4}} \right) \\ \\ \\ = \tan^{-1} \left( \frac{\frac{20+36}{48}}{\frac{48-15}{48}} \right) \\ \\ \\ = \tan^{-1} \left( \frac{56}{33} \right)$

Now, we need to convert it into $\sin^{-1}$ form.

$Let \, \, \tan^{-1} \left( \frac{56}{33} \right) = C \\ \\ \\ \implies \tan C = \frac{56}{33} \\ \\ \\ \implies \cot C = \frac{1}{\tan \, C} = \frac{33}{56} \\ \\ \\ Now \, \, cosec^2 \, C = 1 + \cot^2 C \\ \\ \\ \implies cosec^2 \, C = 1 + \left( \frac{33}{56} \right)^2 \\ \\ \\ \implies cosec^2 \, C = 1 + \frac{1089}{3136} \\ \\ \\ \implies cosec^2 \, C = \frac{3136+1089}{3136} \\ \\ \\ \implies cosec^2 \, C = \frac{4225}{3136} \\ \\ \\ \implies cosec \, C = \frac{65}{56}$


$\implies \sin C = \frac{1}{cosec \, C} = \frac{56}{65} \\ \\ \\ = RHS$

Hence Proved.

Answer 2

Given,
cos^-1 (12/13) + sin^-1 (3/5).
as we know that cos^-1(x)= sin^-1(√1-x²).
.
change' cos^-1 in sin ^-1.
see picture!!