Question

Prove the following​

Answer 1

${\huge{\underline{\underline{\bold{\rm{SolutiOn:}}}}}}$

Taking LHS:-

$2 \sin {}^{ - 1} \frac{3}{5}$

We know that,

$\huge \boxed{2 \sin {}^{ - 1} x = \sin {}^{ - 1} 2x \sqrt{1 - {x}^{2} } }$

$\sin {}^{ - 1} 2 \times \frac{3}{5} \sqrt{1 - (\frac{3}{5}) {}^{2} } \\ = \sin {}^{ - 1} \frac{6}{5} \times \frac{4}{5} \\ = \sin {}^{ - 1} \frac{24}{25}$

Also,

$\huge \boxed{\sin {}^{ - 1} = \tan {}^{ - 1} \frac{x}{ \sqrt{1 - {x}^{2} } } }$

$= \tan {}^{ - 1} \frac{ \frac{24}{25} }{ \sqrt{1 - \frac{576}{625} } } \\ = \tan {}^{ - 1} \frac{ \frac{24}{25} }{ \frac{7}{25} \: } \\ = \tan {}^{ - 1} \frac{24}{7}$

= RHS

______________ᴘʀᴏᴠᴇᴅ.

Answer 2

Answer:

Please see the attachment