Write the value of 2 sin - 1 1 2 + cos - 1 - 1 2 .

Given that, 2 sin - 1 1 2 + cos - 1 - 1 2 = sin - 1 2 × 1 2 1 - 1 2 2 + cos - 1 - 1 2 = sin - 1 3 2 + cos - 1 - 1 2 = sin - 1 sin π 3 + cos - 1 cos 2 π 3 We know that, sin - 1 sin x = x where x ∈ - π 2 , π 2 and cos - 1 cos x = x , where x ∈ 0 , π = π 3 + 2 π 3 = π

Write the value of 2 sin - 1 1 2 + cos - 1 - 1 2 .

Given that, 2 sin - 1 1 2 + cos - 1 - 1 2 = sin - 1 2 × 1 2 1 - 1 2 2 + cos - 1 - 1 2 = sin - 1 3 2 + cos - 1 - 1 2 = sin - 1 sin π 3 + cos - 1 cos 2 π 3 We know that, sin - 1 sin x = x where x ∈ - π 2 , π 2 and cos - 1 cos x = x , where x ∈ 0 , π = π 3 + 2 π 3 = π

Write the value of tan - 1 a b - tan - 1 a - b a + b .

We know that tan - 1 x - tan - 1 y = tan - 1 x - y 1 + x y . Now, tan - 1 a b - tan - 1 a - b a + b = tan - 1 a b - a - b a + b 1 + a b a - b a + b = tan - 1 a 2 + a b - a b + b 2 b ( a + b ) a b + b 2 - a b + a 2 b ( a + b ) = tan - 1 ( 1 ) = tan - 1 tan π 4 ∵ tan π 4 = 1 = π 4 ∴ tan - 1 a b - tan - 1 a - b a + b = π 4

Write the value of cos - 1 cos 5 π 4 .

cos - 1 cos 5 π 4 ≠ 5 π 4 as 5 π 4 does not lie between 0 and π . We have cos - 1 cos 5 π 4 = cos - 1 cos 2 π - 3 π 4 = cos - 1 cos 3 π 4 = 3 π 4

Show that sin - 1 ( 2 x 1 - x 2 ) = 2 sin - 1 x

We have LHS = sin - 1 ( 2 x 1 - x 2 ) Putting x = sin a , we get = sin - 1 ( 2 sin a 1 - sin 2 a ) = sin - 1 ( 2 sin a cos a ) = sin - 1 ( sin 2 a ) = 2 a = 2 sin - 1 x ( ∵ x = sin a )

Evaluate: sin - 1 sin 3 π 5

We know that sin - 1 ( sin x ) = x We have sin - 1 sin 3 π 5 = sin - 1 sin π - 3 π 5 ∵ π - 3 π 5 ∈ - π 2 , π 2 = sin - 1 sin 2 π 5 = 2 π 5 ∴ sin - 1 sin 3 π 5 = 2 π 5