Question

prove that cos^-1(63/65) + 2 tan^-1(1/5) = sin^-1(3/5)

Answer 1

The solution is attached as a word file

684c21e9102279e0ea4f0ffa5fd03da9.doc

Answer 2

Answer:

Proof below.

Step-by-step explanation:

Given : $\cos^{-1}(\frac{63}{65}) + 2 \tan^{-1}(\frac{1}{5}) = \sin^{-1}(\frac{3}{5})$

To prove : The given expression?

Solution :

LHS - $\cos^{-1}(\frac{63}{65}) + 2 \tan^{-1}(\frac{1}{5})$

We know, Using the identity

$\cos^{-1}(\frac{a}{b})=\tan^{-1}(\frac{\sqrt{b^2-a^2}}{a})$

$2\tan^{-1}a=\tan^{-1}(\frac{2a}{1-a^2})$

$\tan^{-1}a+\tan^{-1}b=\tan^{-1}(\frac{a+b}{1-ab})$

$\tana{-1}(\frac{a}{b})=\sin^{-1}(\frac{a}{\sqrt{a^2+b^2}})$

Substitute in,

$\cos^{-1}(\frac{63}{65}) =\tan^{-1}(\frac{\sqrt{65^2-63^2}}{63})=\tan^{-1}(\frac{16}{63})$

$2 \tan^{-1}(\frac{1}{5})=\tan^{-1}(\frac{2(\frac{1}{5})}{1-(\frac{1}{5})^2})=\tan^{-1}(\frac{5}{12})$

Substitute in LHS,

$=\cos^{-1}(\frac{63}{65}) + 2 \tan^{-1}(\frac{1}{5})$

$=\tan^{-1}(\frac{16}{63}) +\tan^{-1}(\frac{5}{12})$

$=\tan^{-1}(\frac{\frac{16}{63}+\frac{5}{12}}{1-(\frac{16}{63})(\frac{5}{12})})$

$=\tan^{-1}(\frac{3}{4})$

$=\sin^{-1}(\frac{3}{\sqrt{3^2+4^2}})$

$=\sin^{-1}(\frac{3}{5})$

$=RHS$

Therefore, LHS=RHS