Test: Introduction to Inverse Trigonometry - JEE MCQ

# Test: Introduction to Inverse Trigonometry - JEE MCQ

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## 10 Questions MCQ Test Mathematics (Maths) for JEE Main & Advanced - Test: Introduction to Inverse Trigonometry

Test: Introduction to Inverse Trigonometry for JEE 2024 is part of Mathematics (Maths) for JEE Main & Advanced preparation. The Test: Introduction to Inverse Trigonometry questions and answers have been prepared according to the JEE exam syllabus.The Test: Introduction to Inverse Trigonometry MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Introduction to Inverse Trigonometry below.
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Test: Introduction to Inverse Trigonometry - Question 1

### The value of  is given by​

Detailed Solution for Test: Introduction to Inverse Trigonometry - Question 1

Test: Introduction to Inverse Trigonometry - Question 2

### The principal value of tan-1 (-1) is given by​:

Detailed Solution for Test: Introduction to Inverse Trigonometry - Question 2

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​

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Test: Introduction to Inverse Trigonometry - Question 3

### if 5 sin θ = 3, then  is equal to

Detailed Solution for Test: Introduction to Inverse Trigonometry - Question 3

sin θ is 3/5.
on simplifying:
(secθ + tanθ)/(secθ - tanθ)
We get, (1+sin θ)/(1-sin θ)
=(1+3/5)/(1-3/5)
=(8/2)
=4

Test: Introduction to Inverse Trigonometry - Question 4

Value of

Detailed Solution for Test: Introduction to Inverse Trigonometry - Question 4

Since sin-1(-π/2) = -sin-1(π/2) = -1

so, cot -1(-1)

let (- cot-11) = α

cot α = -1 = cot (π/2+π/4) = 3π/4

Test: Introduction to Inverse Trigonometry - Question 5

What is the solution of cot⁡(sin-1⁡x)?

Detailed Solution for Test: Introduction to Inverse Trigonometry - Question 5

Let sin-1⁡x = y.
From ∆ABC, we get

y = sin-1⁡x

∴cot⁡(sin-1⁡x)

Test: Introduction to Inverse Trigonometry - Question 6

The principal value of
is.

Detailed Solution for Test: Introduction to Inverse Trigonometry - Question 6

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

Test: Introduction to Inverse Trigonometry - Question 7

If x< 0 then value of tan-1(x) + tan-1 (1/x) is equal to

Detailed Solution for Test: Introduction to Inverse Trigonometry - Question 7

We know that

Test: Introduction to Inverse Trigonometry - Question 8

Find the value of tan-1⁡(1/3) + tan-1⁡(1/5) + tan-1⁡(1/7).

Detailed Solution for Test: Introduction to Inverse Trigonometry - Question 8

Test: Introduction to Inverse Trigonometry - Question 9

Evaluate

Detailed Solution for Test: Introduction to Inverse Trigonometry - Question 9

Test: Introduction to Inverse Trigonometry - Question 10

What is the value of 2 tan-1⁡x?

Detailed Solution for Test: Introduction to Inverse Trigonometry - Question 10

Let 2 tan-1⁡x = y

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## Mathematics (Maths) for JEE Main & Advanced

209 videos|443 docs|143 tests