Important Questions: Inverse Trigonometric Functions | Mathematics (Maths) Class 12 - JEE PDF Download

Q1: Find the value of cos^-1 (1/2) + 2 sin^-1 (1/2).
Ans: First, solve for cos^-1 (1/2):
Let us take, y = cos^-1 (1/2)
This can be written as:
cos y = (1/2)
cos y = cos (π /3).
Thus, the range of principal value of cos^-1 is [0, π ]
Therefore, the principal value of cos^-1 (1/2) is π/3.
Now, solve for sin^-1 (1/2):
Let y = sin^-1 (1/2)
sin y = 1/2
sin y = sin ( π/6)
Thus, the range of principal value of sin^-1 is [(-π)/2, π/2 ]
Hence, the principal value of  sin^-1 (1/2) is π/6.
Now we have cos^-1 (1/2) = π/3 & sin^-1 (1/2) = π/6
Now, substitute the obtained values in the given formula, we get:
= cos^-1 (1/2) + 2sin^-1 (1/2)
= π /3 + 2( π/6)
= π/3 + π/3
= ( π+π )/3
= 2π /3
Thus, the value of cos^-1 (1/2) + 2 sin^-1 (1/2) is  2π /3.

Q2: The value of tan^-1 √3 – sec^-1(–2) is equal to:
Ans: Now, solve the first part of the expression:  tan^-1 √3
Let us take y = tan^-1√3
This can be written as:
tan y = √3
Now, use the trigonometry table to find the radian value
tan y = tan (π/3)
Thus, the range of principal value of tan^-1 is (−π/2, π/2)
Therefore, the principal value of tan^-1√3 is π/3.
Now, solve the second part of the expression: sec^-1(–2)
Now, assume that y = sec^-1 (–2)
sec y = -2
sec y = sec (2π/3)
We know that the principal value range of sec^-1 is [0,π] – {π/2}
Therefore, the principal value of sec^-1 (–2) = 2π/3
Now we have:
tan^-1(√3) = π/3
sec^-1 (–2) = 2π/3
Now, substitute the values in the given expression:
= tan^-1 √3 – sec^-1 (−2)
= π/3 − (2π/3)
= π/3 − 2π/3
= (π − 2π)/3
= – π/3

Q3: Determine the principal value of cos^-1( -1/2).
Ans: Let us assume that, y = cos^-1( -1/2)
We can write this as:
cos y = - 1/2
cos y = cos (2π/3).
Thus, the Range of the principal value of cos^-1 is [0, π ].
Therefore, the principal value of  cos^-1( -1/2) is 2π /3.

Q4: Prove that sin^-1 (3/5) – sin^-1 (8/17) = cos^-1 (84/85).
Ans: Let sin^-1 (3/5) = a and sin^-1 (8/17) = b
Thus, we can write sin a = 3/5 and sin b = 8/17
Now, find the value of cos a and cos b
To find cos a:
Cos a = √[1 – sin² a]
= √[1 – (3/5)² ]
= √[1 – (9/25)]
= √[(25-9)/25]
= 4/5
Thus, the value of cos a = 4/5
To find cos b:
Cos b = √[1 – sin² b]
= √[1 – (8/17)² ]
= √[1 – (64/289)]
= √[(289-64)/289]
= 15/17
Thus, the value of cos b = 15/17
We know that cos (a- b) = cos a cos b + sin a sin b
Now, substitute the values for cos a, cos b, sin a and sin b in the formula, we get:
cos (a – b) = (4/5) x (15/17) + (3/5) x (8/17)
cos (a – b) = (60 + 24)/(17x 5)
cos (a – b) = 84/85
(a – b) = cos^-1 (84/85)
Substituting the values of a and b sin^-1 (3/5)- sin^-1 (8/7) = cos^-1 (84/85)
Hence proved.

Q5: Find the value of cot (tan^-1 α + cot^-1 α).
Ans: Given that: cot (tan^-1 𝛂 + cot^-1 𝛂)
= cot (𝝅/𝟐) (since, tan^-1 x + cot^-1 x = 𝜋/2)
= cot (180°/2) ( we know that cot 90° = 0)
= cot (90°)
= 0
Therefore, the value of cot (tan^-1 α + cot^-1 α) is 0.