Question

Prove it,


$\bf \: {sin}^{ - 1} \frac{3}{5} - {cos}^{ - 1} \frac{63}{65} = 2 \: tan {}^{ - 1} \frac{1}{5}$
​

Answer 1

Step-by-step explanation:

Given Equation is sin⁻¹(3/5) - cos⁻¹(63/65) = 2tan⁻¹(1/5).

It can be written as,

⇒ -cos⁻¹(3/5) - 2tan⁻¹(1/5) = -sin⁻¹(3/5)

⇒ cos⁻¹(3/5) + 2tan⁻¹(1/5) = sin⁻¹(3/5)

$\textbf{\underline{\underline{Important Formulas}}}$

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

$\boxed{sin^{-1}x + sin^{-1}y = sin^{-1}[(x\sqrt{1-x^2}] + y[\sqrt{1-x^2}]}$

LHS:

$=sin^{-1}\sqrt{1-(\frac{63}{65})^2} + sin^{-1} \frac{2 * \frac{1}{5}}{1 + (\frac{1}{5})^2}$

$=sin^{-1}\sqrt{1-\frac{3969}{4225}} + sin^{-1}\frac{\frac{2}{5}}{\frac{26}{25}}$

$=sin^{-1}\sqrt{\frac{256}{4225}} + sin^{-1}\frac{5}{13}$

$=sin^{-1}\frac{16}{65} + sin^{-1}\frac{3}{5}$

$=sin^{-1}[(\frac{16}{65}\sqrt{1-(\frac{5}{13})^2} + \frac{5}{13}\sqrt{1-(\frac{16}{65})^2}]$

$=sin^{-1}(\frac{16}{65} * \frac{12}{13} + \frac{5}{13} * \frac{63}{65})$

$=sin^{-1}(\frac{16*12+315}{13*65})$

$=sin^{-1}(\frac{507}{845})$

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

RHS

Hope it helps!

Answer 2

Answer:

Step-by-step explanation:

LHS:

=sin^{-1}\sqrt{1-(\frac{63}{65})^2} + sin^{-1} \frac{2 * \frac{1}{5}}{1 + (\frac{1}{5})^2}

=sin^{-1}\sqrt{1-\frac{3969}{4225}} + sin^{-1}\frac{\frac{2}{5}}{\frac{26}{25}}

=sin^{-1}\sqrt{\frac{256}{4225}} + sin^{-1}\frac{5}{13}

=sin^{-1}\frac{16}{65} + sin^{-1}\frac{3}{5}

=sin^{-1}[(\frac{16}{65}\sqrt{1-(\frac{5}{13})^2} + \frac{5}{13}\sqrt{1-(\frac{16}{65})^2}]

=sin^{-1}(\frac{16}{65} * \frac{12}{13} + \frac{5}{13} * \frac{63}{65})

=sin^{-1}(\frac{16*12+315}{13*65})

=sin^{-1}(\frac{507}{845})

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

RHS

Hope it helps!