Question

sin⁻¹ 3/5 + sin⁻¹ 8/17 + sin⁻¹ 36/85 =π/2,Prove it.

Answer 1

we have to prove : sin^-1(3/5) + sin^-1(8/17) + sin^-1(36/85) = π/2

LHS = sin^-1(3/5) + sin^-1(8/17) + sin^-1(36/85)

= [sin^-1(3/5) + sin^-1(36/85)] + sin^-1(8/17)

we know, sin^-1x + sin^-1y = sin^-1(x√(1-y^2) + y√(1 - x^2)]

so, sin^-1(3/5) + sin^-1(36/85) = sin^-1[3/5 × √(1 - 36²/85²) + 36/85 × √(1 - 3²/5²) ]

= sin^-1[3/5 × 77/85 + 36/85 × 4/5]

[ because √(1 - 36²/85²) = 77/85 and √(1 - 3²/5²) = 4/5]

= sin^-1[231/425 + 144/425]

= sin^-1[(231 + 144)/425]

= sin^-1[375/425]

= sin^-1(15/17)

also, sin^-1(15/17) = cos^-1(8/17)
hence, [sin^-1(3/5) + sin^-1(36/85)] = cos^-1(8/17)

now, [sin^-1(3/5) + sin^-1(36/85)] + sin^-1(8/17)
= cos^-1(8/17) + sin^-1(8/17)

we know, sin^-1x + cos^-1x = π/2 for all x belongs to -1 ≤ x ≤ 1

so, cos^-1(8/17) + sin^-1(8/17) = π/2 = RHS

Answer 2

HELLO DEAR,




GIVEN:-
sin-¹3/5 + Sin-¹8/17 + sin-¹ 36/85 = π/2


we know:- Sin-¹x + Sin-¹y = Sin-¹[x√(1 - y²) + y√(1 - x²)]

so, (sin-¹3/5 + Sin-¹8/17) + sin-¹ 36/85

=> sin-¹ [(3/5)√(1 - 8²/17²) + (8/17)√(1 - 3²/5²)] + sin-¹ 36/85

=> sin-¹ [(3/5)(15/17) + (8/17)(4/5)] + sin-¹36/85

=> sin-¹ [9/17 + 32/85] + sin-¹ 36/85

=> sin-¹ (45 + 32)/85 + sin-¹ 36/85

=> sin-¹ 77/85 + sin-¹ 36/85

let sin-¹ 77/85 = A so, sinA = 77/55
cosA = √(1 - sin²A)
cosA = √{(7225 - 5928)/7225}
cosA = √1296/7225
cosA = 36/85
cos-¹ 36/85 = A = sin-¹ 77/85

therefore,
cos-¹36/85 + sin-¹ 36/85 = π/2

[as we know:- sin-¹x + cos-¹x = π/2]

hence, proved,


I HOPE ITS HELP YOU DEAR,
THANKS