Question

Prove that cos⁻¹ (12/13) + sin⁻¹ (3/5) = sin⁻¹ (56/65)

Answer 1

$\text{L.H.S}\\\\=\cos^{-1} \dfrac{12}{13} + \sin^{-1} \dfrac 35\\\\=\sin^{-1} \dfrac 5{13} + \sin^{-1} \dfrac 35$

$=\sin^{-1}\left[\dfrac 5{13}\sqrt{1- \left(\dfrac 35 \right)^2} + \dfrac 35\sqrt{1-\left(\dfrac 5{13} \right)^2} \right]\\\\=\sin^{-1} \left(\dfrac 5{13} \sqrt{1-\dfrac 9{25} }+\dfrac 35 \sqrt{1-\dfrac{25}{169}} \right)\\\\=\sin^{-1} \left(\dfrac 5{13} \sqrt{\dfrac{16}{25}}+\dfrac 35 \sqrt{\dfrac{144}{169}} \right)\\\\=\sin^{-1} \left(\dfrac{5}{13} \cdot \dfrac 45 + \dfrac 35 \cdot \dfrac{12}{13} \right)$

$=\sin^{-1} \left(\dfrac 4{13} +\dfrac{36}{65}\right)\\\\=\sin^{-1} \left(\dfrac{20}{65} + \dfrac{36}{65} \right)\\\\=\sin^{-1} \left(\dfrac{20+36}{65} \right)\\\\=\sin^{-1} \left(\dfrac{56}{65} \right)\\\\=\text{R.H.S}$

Answer 2

Answer:

See below ↓

Step-by-step explanation:

We need to prove :

⇒ cos⁻¹ $\frac{12}{13}$ + sin⁻¹ $\frac{3}{5}$ = tan⁻¹ $\frac{56}{65}$

Let's simplify the LHS.

• cos⁻¹ $\frac{12}{13}$ + sin⁻¹ $\frac{3}{5}$

Convert the inverse cos and sin functions into inverse tan functions

• tan⁻¹ $\frac{5}{12}$ + tan⁻¹ $\frac{3}{4}$
• [∴This can be found taking a right triangle and labeling the sides,   and then using Pythagorean Theorem, we can find the missing side and take the ratio of tan]

Identity

• tan⁻¹ x + tan⁻¹ y = tan⁻¹ $\frac{x+y}{1-xy}$

Using this identity, we can simplify our earlier equation!

⇒ tan⁻¹ [(5/12 + 3/4)/(1 - (5/12 x 3/4))]

⇒ tan⁻¹ [(20 + 36) / (48 - 15)

⇒ tan⁻¹ (56/65)

⇒ RHS

⇒ Proved ∴√