Question

sin⁻¹ 12/13 + cos⁻¹ 4/5 + tan⁻¹ 63/16=π,Prove it.

Answer 1

we have to prove that,

sin⁻¹ (12/13) + cos⁻¹ (4/5) + tan⁻¹ (63/16)=π

Let sin^-1(12/13) = A

⇒sinA = 12/13

then, tanA = 12/5

similarly, cos^-1(4/5) = B

⇒cosB = 4/5

then, tanB = 3/4

we know,

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

= (12/5 + 3/4)/(1 - 12/5 × 3/4)

= (48 + 15)/(5 × 4 - 12 × 3)

= 63/-16

hence, tan(A + B) = -63/16

⇒A + B = tan^-1(-63/16)

⇒sin⁻¹ (12/13)+ cos⁻¹ (4/5) = tan^-1(-63/16)

now, sin⁻¹ 12/13 + cos⁻¹ 4/5 + tan⁻¹(63/16)

= tan^-1(-63/16) + tan^-1(63/16)

use formula,

tan^-1x + tan^-1y = tan^-1(x + y)/(1 - xy)

so, tan^-1(-63/16) + tan^-1(63/16) = tan^-1(-63/16 + 63/16)/{1 - (-63/16) × (63/16)}

= tan^-1(0)

= π

hence, sin⁻¹ 12/13 + cos⁻¹ 4/5 + tan⁻¹ 63/16=π