Question

Prove it,


$\bf \: {cos}^{ - 1} \frac{3}{5} + {cos}^{ - 1} \frac{12}{13} = {sin}^{ - 1} \frac{63}{65}$
​

Answer 1

HELLO DEAR,

we know:- $\bf{cos^{-1}x + cos^{-1}y = cos^{-1}\{{xy - \sqrt{(1 - x^2)(1 - y^2)}}\}}$

now,

$\bf{cos^{-1}3/5 + cos^{-1}12/13}$

$\bf{\implies cos^{-1}\{(3/5)(12/13) - \sqrt{(1 - 9/25)(1 - 144/169)}\}}\\ \\ \bf{\implies cos^{-1}\{36/65 - \sqrt{(16/25)(25/169)}\}} \\\\ \bf{\implies cos^{-1}\{36/65 - 4/13\}}\\ \\ \bf{\implies cos^{-1} \{\frac{36 - 20}{65}\}}\\ \\\bf{\implies cos^{-1} \frac{16}{65}}$

we also know that :- $\bold{cos^{-1}z = sin^{-1}\sqrt{1 - z^2}}$

$\bold{\therefore , cos^{-1}16/65 = sin^{-1}\sqrt{1 - 256/4225}}$

$\bold{\implies sin^{-1}\sqrt{\frac{4225 - 256}{4225}} = sin^{-1}63/65}$

$\bold{\implies cos^{-1}3/5 + cos^{-1}12/13 = sin^{-1}63/65}$

I HOPE IT'S HELP YOU DEAR,
THANKS

Answer 2

Answer:

HELLO DEAR,

we know:- \bf{cos^{-1}x + cos^{-1}y = cos^{-1}\{{xy - \sqrt{(1 - x^2)(1 - y^2)}}\}}cos

−1

x+cos

−1

y=cos

−1

{xy−

(1−x

2

)(1−y

2

)

}

now,

\bf{cos^{-1}3/5 + cos^{-1}12/13}cos

−1

3/5+cos

−1

12/13

\begin{lgathered}\bf{\implies cos^{-1}\{(3/5)(12/13) - \sqrt{(1 - 9/25)(1 - 144/169)}\}}\\ \\ \bf{\implies cos^{-1}\{36/65 - \sqrt{(16/25)(25/169)}\}} \\\\ \bf{\implies cos^{-1}\{36/65 - 4/13\}}\\ \\ \bf{\implies cos^{-1} \{\frac{36 - 20}{65}\}}\\ \\\bf{\implies cos^{-1} \frac{16}{65}}\end{lgathered}

⟹cos

−1

{(3/5)(12/13)−

(1−9/25)(1−144/169)

}

⟹cos

−1

{36/65−

(16/25)(25/169)

}

⟹cos

−1

{36/65−4/13}

⟹cos

−1

{

65

36−20

}

⟹cos

−1

65

16

we also know that :- \bold{cos^{-1}z = sin^{-1}\sqrt{1 - z^2}}cos

−1

z=sin

−1

1−z

2

\bold{\therefore , cos^{-1}16/65 = sin^{-1}\sqrt{1 - 256/4225}}∴,cos

−1

16/65=sin

−1

1−256/4225

\bold{\implies sin^{-1}\sqrt{\frac{4225 - 256}{4225}} = sin^{-1}63/65}⟹sin

−1

4225

4225−256

=sin

−1

63/65

\bold{\implies cos^{-1}3/5 + cos^{-1}12/13 = sin^{-1}63/65}⟹cos

−1

3/5+cos

−1

12/13=sin

−1

63/65

I HOPE IT'S HELP YOU DEAR,

THANKS