Inverse Trigonometric Functions Test - 3

Let y = tan^-1 (-1)

tan y = -1

tan y = tan(-π/4)

Range of principal value of tan^-1   is
(-π/2 , π/2)

SInce - 1 is negative,
the principle value of tan^-1 (-1) is - π/4 ​

sin θis 3/5.
on simplifying:
(sec θ + tan θ)/(sec θ- tan θ)
We get, (1+sin θ)/(1-sin θ)
=(1+3/5)/(1-3/5)
=(8/2)
=4

Since sin^-1 (-π/2) = -sin^-1 (π/2) = -1

so, cot ^-1 (-1)

let (- cot^-1 1) = α

cot α= -1 = cot (π/2+π/4) = 3 π/4

Let sin^-1 ⁡x = y.
From ∆ABC, we get

y = sin^-1 ⁡x
= 

∴cot ⁡(sin^-1 ⁡x)

=  

tan^-1  (tan 3 π/5)
This can be written as:
tan^-1  (tan 3 π/5) = tan^-1  (tan[π–2 π/5])
= tan^-1  (- tan 2 π/5) {since tan(π–x) = -tan x}
= –tan^-1  (tan 2 π/2)
= –2 π/5

We know that

 Let 2 tan^-1 ⁡x = y