Chapter 2: Inverse Trigonometric Functions

We know that,

tan^-1 tan x = x, x ∈ (-π/2, π/2)

And, here

tan^-1 tan (5π/6) ≠ 5π/6 as 5π/6 ∉ (-π/2, π/2)

Also,

cos^-1 cos x = x; x ∈ [0, π]

So,

cos^-1 cos (13π/6) ≠ 13π/6 as 13π/6 ∉ [0, π]

Now,

tan^-1 [tan (5π/6)] + cos^-1 [cos (13π/6)]

= tan^-1 [tan (π – π /6)] + cos^-1 [cos (2π + π/6)]

= tan^-1 [ -tan π /6] + cos^-1 [ -cos (7π/6)]

= – tan^-1 [tan π /6] + cos^-1 [cos (π/6)]

= – π /6 + π /6

= 0

Re-writing the given,

L.H.S = R.H.S

– Hence Proved

Given,

tan^-1 (-1/√3) + cot^-1(1/√3) + tan^-1(sin (-π/2))

We know that,

tan^-1 tan x = x, x ∈ (-π/2, π/2)

Taking L.H.S = 2 tan^-1(-3) = -2 tan^-1 3 (∵ tan^-1 (-x) = – tan^-1 x, x ∈ R)

= R.H.S

– Hence Proved.

Given equation,

Hence, the real solutions of the given trigonometric equation are 0 and -1.

Given expression, sin (2 tan^-1 1/3) + cos (tan^-1 2√2)

Given, 2 tan^-1(cos θ) = tan^-1 (2 cosec θ)

So,

Taking L.H.S, we have

Thus, L.H.S = R.H.S

– Hence proved