Solved Examples for JEE: Inverse Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced PDF Download

Equations Involving Inverse Trigonometric Functions 

Ex.1 Solve 

Sol. 

Given:

  ...(1)

or cos^–1 x√3 = π/2 – cos^–1 x

or cos cos^–1 x√3 = cos (π/2 – cos^–1x)

or x√3 = sin cos^–1x    or

or x√3 =  

Squaring we get:

3x² = 1 – x²  or 4x² = 1 = x = ± 1/2

Verification : When x = 1/2

L.H.S. of equation = cos1 ( 3 /2) + cos^–1 (1/2) = π/6 + π/3 +π/2 = R.H.S. of equation

When x = –1/2

L.H.S. of equation =  cos^–1 (– 3 /2) + cos^–1 (–1/2) = π – cos^–1 ( 3 /2) + π – cos^–1 (1/2)

= π – π/6 + π – π/3 = 3p/2 ≠ R.H.S. of equation

∴ x = 1/2 is the only solution

Ex.2 Solve for x : (tan^-1 x)² + (cot^-1 x)² = 

Sol. 

 tan^–1 x = - π/4, 3 π/4  =  tan^–1 x = – π/4;  x = –1

Ex.3 Determine the integral values of ' k ' for which the system , (arc tan x)² + (arc cos y)² = π² k and tan ^-1 x + cos ^-1 y = π /2 posses solution and find all the solutions. 

Sol.

  = 1 - 2 + 8 k ≥ 0 = k ≥ 1/2 ..(2)

From  (1)  and  (2)    k = 1 

Inequations involving inverse trigonometric functions 

Ex.1 Find the interval in which cos^-1 x > sin^-1 x.

Sol. 

We have, cos^–1 x > sin^–1 {for cos–1 x to be real; x E [–1, 1]}

2 cos^–1 x > π/2  = cos^–1 x > π/4   or  cos (cos^–1 x) < cos π/4

Ex.2 Find the solution set of the inequation sin^-1(sin 5) > x² - 4x

Sol. 

sin^–1(sin 5) > x² – 4x ⇒ sin–1[sin(5 – 2π)] > x² – 4x

⇒ x² – 4x < 5 – 2π ⇒ x² – 4x + (2π – 5) < 0

Summation of Series 

Ex.1 Sum the series , 

Sol. 

= tan ^-1 (n + 1) (n + 2) - tan ^-1 n (n + 1)

Put n  =  1 , 2 , 3 , ........ , n and add, we get S_n  = tan ^-1 (n + 1) (n + 2)  -  tan ^-1 2

Ex.2 Sum the series to ' n ' terms , + ...... to ' n ' terms. Also show that , S_∞  = tan ^-1 3 .

Sol. 

= tan ^-1 (n + 2)  –  tan ^-1 (n)

Hence, S_n = tan ^-1 (n + 2)  +  tan ^-1 (n + 1) - (tan ^-1 1  +  tan ^-1 2)

Ex.3 If the sum , find the value of k.

Sol.

 .....