Question

Prove that:(its urgent plz help someone who's intelligent enough !!)
$sin^-^1 \frac{4}{5} +sin^-^1 \frac{5}{13} +sin^-^1 \frac{16}{65} = \frac{ \pi }{2}$

Answer 1

sin-1(x) + sin-1(y) =sin-1((x*sqrt(1-y²) + y*sqrt(1-x²))

sin-1(4/5) + sin-1(5/13) =sin-1((4/5*sqrt(1-25/169) + 5/13*sqrt(1-16/25))

sin-1((4/5*sqrt(144/169) + 5/13*sqrt(9/25))

sin-1((4/5)*(12/13) + (5/13)*3/5))

sin-1((48/65) + 15/65))

sin-1(63/65)

sin-1(63/65) + sin-1(16/65) =sin-1((63/65*sqrt(1-256/4225) + 16/65*sqrt(1-3969/4225))

sin-1((63/65*sqrt(3969/4225) + 16/65*sqrt(256/4225))

sin-1((63/65)*(63/65) + (16/65)*16/65))

sin-1(65²/65²)

sin-1(1) =  π/2

Answer 2

Let  Sin⁻¹ 4/5 = A      =>  Sin A = 4/5       =>  Cos A = √(1 - 4²/5²) = 3/5

Let  Sin⁻¹ 5/13 = B      => Sin B = 5/13       =>  Cos B = 12/13

Let  Sin⁻¹ 16/65 = C      =>  Sin C = 16/65     =>  Cos C = √(1 - 16²/65²) = 63/65
   
LHS = A + B + C.

Let us find:
   Sin (B+C) = Sin B Cos C + Cos B Sin C
                  = 5/13 * 63/65 + 12/13 * 16/65 = 0.6
   thus  Cos (B+C) = √(1 - 0.6²) = 0.8

Finding Sine of LHS :
        Sin [A + (B+C)] = Sin A Cos(B+C)  + Cos A Sin (B+C)
               = 4/5 * 0.8 + 3/5 * 0.6 = 1

Hence,  A+B+C = π/2