Question

cos⁻¹ 4/5 + sin⁻¹ 5/13 = tan⁻¹(56/33),Prove it.

Answer 1

we have to prove that,

cos⁻¹ 4/5 + sin⁻¹ 5/13 = tan⁻¹(56/33)

LHS = cos⁻¹ 4/5 + sin⁻¹ 5/13

Let cos^-1(4/5) = A ⇒cosA = 4/5

then, tanA = √(5² - 4²)/4 = 3/4.....(1)

similarly, sin^-1(5/13) = B ⇒sinB = 5/13

then, tanB = 5/√(13² - 12²) = 5/12 .....(2)

so, cos⁻¹ 4/5 + sin⁻¹ 5/13 = A + B ........(3)

we know,

tan(A + B) = (tanA + tanB)/(1 - tanA.tanB)

from equations (1) and (2),

= (3/4 + 5/12)/(1 - 3/4 × 5/12)

= (36 + 20)/(4 × 12 - 3 × 5)

= 56/(48 - 15)

= 56/33

so, A + B = tan^-1(56/33) = RHS .......(4)

from equations (3) and (4),

cos⁻¹ (4/5) + sin⁻¹ (5/13) = tan⁻¹(56/33)

Answer 2

Answer with Explanation:

---------------------[To prove ]----------

$\sf cos^{-1}\frac{4}{5}+sin^{-1}\frac{5}{13}=tan^{-1}(\frac{56}{33})$

---------------------[proof ]---------------

Let,

$\sf cos^{-1}\frac{4}{5}=x$

$\implies \sf cosx =\frac{4}{5}$

we know, sin²Φ+cos²Φ = 1 ,

then, sin²Φ = 1-cos²Φ

$\implies \sf sinx =\sqrt{1-cos^2x}=\sqrt{1-(\frac{4}{5})^2}$

$\implies \sf \sqrt{1-(\frac{16}{25})}=\frac{3}{5}$

therefore,

$\sf tanx=\frac{sinx}{cosx}=\frac{3}{4}$--------①

Now, Let sin⁻¹ 5/13 = y

therefore, $\sf siny =\frac{5}{13}$

$\sf cosy =\sqrt{1-sin^2y}=\sqrt{1-(\frac{5}{13})^2}$

$\sf cosy =\sqrt{1-(\frac{25}{169})}=\sqrt{144}{169}=\frac{12}{13}$

So, $\sf tany =\frac{5}{12}$--------②

we know,

$\sf tan(x+y)=\frac{tanx+tany}{1-tanx.tany}$

$\implies \sf \frac{\frac{3}{4}+\frac{5}{12}}{1-(\frac{3}{4})(\frac{5}{12})}$

$\implies \sf \frac{56/48}{33/48}=\frac{56}{33}$

tan(x+y)=56/33 - - - - - - -③

from equation ① , ② , ③

$\sf cos^{-1}\frac{4}{5}+sin^{-1}\frac{5}{13}=tan^{-1}(\frac{56}{33})$