Prove that:(its urgent plz help someone who's intelligent enough !!)

2
by vivek2001

2015-02-17T20:56:57+05:30

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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
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haahaaa u r pravin undoubtedly.........yeh to isliye likha tha ki question catchy lage
thank uuuuuuuuuuuuuuuuuuuu
Mention not
2015-02-18T12:04:56+05:30

Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
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

this is a solution , in a simple way. i hope u r able to understand