Important Questions: Inverse Trigonometric Functions

# 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.

The document Important Questions: Inverse Trigonometric Functions | Mathematics (Maths) Class 12 - JEE is a part of the JEE Course Mathematics (Maths) Class 12.
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## Mathematics (Maths) Class 12

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## FAQs on Important Questions: Inverse Trigonometric Functions - Mathematics (Maths) Class 12 - JEE

 1. What are inverse trigonometric functions?
Ans. Inverse trigonometric functions are mathematical functions that provide the angle or angles whose trigonometric ratios match a given value. They are used to find the angle when the ratio of sides in a right triangle is known.
 2. How do inverse trigonometric functions relate to trigonometric functions?
Ans. Inverse trigonometric functions are the inverse operations of the corresponding trigonometric functions. They help in finding the angle that produces a specific trigonometric ratio, whereas trigonometric functions provide the ratio of sides given an angle.
 3. What are the commonly used inverse trigonometric functions?
Ans. The commonly used inverse trigonometric functions are arcsine (sin⁻¹), arccosine (cos⁻¹), and arctangent (tan⁻¹). These functions are denoted by placing an arc in front of the trigonometric function.
 4. How are inverse trigonometric functions useful in real-life applications?
Ans. Inverse trigonometric functions have various applications in real-life scenarios, such as in physics, engineering, and navigation. They are used to calculate angles for satellite dish alignment, determining heights and distances, analyzing pendulum motion, and designing roller coasters, among others.
 5. How can inverse trigonometric functions be evaluated?
Ans. Inverse trigonometric functions can be evaluated using calculators or mathematical tables. They can also be calculated using special formulas or by applying the properties and definitions of trigonometric functions. Additionally, inverse trigonometric functions can be represented graphically on the unit circle.

## Mathematics (Maths) Class 12

204 videos|288 docs|139 tests

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