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Trigonometry 3b ( Solved Problems and companion MCQ Tests related to Inverse Trigonometric Ratios )

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                                                   Problems and Quizzes related to Inverse Trigonometric Ratios

Inverse Trigonometric Ratios

Target Audience: High School Students, College Freshmen and Sophomores, Class 11/12 Students in India preparing for ISC/CBSE and Entrance Examinations like the IIT-JEE, Anyone else who needs this Tutorial as a reference!

Solved Problems:
1. Problem : Find the principal value cot-1(-13)
2. Problem : Evaluate each of the following:
(a) cos (Arcsin 3/5),
(b) sin [Arccos (-2/3)],
(c) tan [Arcsin (-3/4)]
3. Problem : Show that
(i) sin-1 ( 2x1-x2) = 2 sin-1x, -12≤x≤ 12
(ii) sin-1 (2x1-x2) = 2 cos-1x,  12≤x≤1
4. Problem : Find the domain of sin-12x+π/6.

5. Problem : Express tan-1( cosx1-sinx ), -π2<x<π2 in the simplest form.

6. Problem : Find the value of sin-1 (sin 3π/5)
7. Problem : Show that sin-112/13 + cos-14/5 + tan-163/16 = π
8. Problem : Evaluate cos (Arctan 15/8 - Arcsin 7/25).
9. Problem : Solve tan-12x + tan-13x = π/4
10. Problem : Prove that 2 tan-113 + tan-117 = π/4

11. Problem : Solve the equation

tan-1x+1x-1 + tan-1x-1x = tan-1(-7)

Solution : Here we have

a + b = c


a = tan-1x+1x-1  => tana = x+1x-1

b = tan-1x-1x  => tanb = x-1x


c = tan-1(-7) => tanc = -7

Since tan(a+b) = tanc
tana+tanb1-tana.tanb = tanc
i. e.     x+1x-1+x-1x1-x+1x-1.x-1x = -7
i.e.     2x2-x+11-x = -7

so that x=2

This value makes the left-hand side of the given equation positive, so  that there is no value of x strictly satisfying the given equation.

The value x = 2 is a solution of the equation
tan-1x+1x-1 + tan-1x-1x = π + tan-1(-7)

12. Problem : Solve Arccos 2x - Arccos x = π/3 .
13. Problem : Graph
i) y = cos-1x + 1 ii)  y = sin-1(x-2)

14. Problem : Show that one can use a composition of trigonometry buttons such as, sin, cos, tan, sin−1, cos−1, and tan−1, to replace the broken reciprocal button on a
15. Problem : A camera is placed on a deck of a pool. A diver is 18 feet above the camera lens. The extended length of the diver is 8 feet.

16. Problem : If sin-1 x + (sin-1 y + sin-1 z) = π/2, Find out x2 + y2 + z2 + 2xyz. 

17. Problem : If  sin-1(x-  x22 +  x34 ....) cos-1(x2 -  x42 +  x64 - ....) = π/2
18. Problem : Solve the equation:

tan-1 2x + tan-1 3x = nπ + π/4

19. Problem : Which angle is greater?

A = 2tan-1(22 - 1) and B = 3sin-1(13) + sin-1(35)

Solution : We observe 22 - 1 2(1.4) – 1 = 2.8 – 1 = 1.8
So 22 – 1 > 3 => 2tan-1(22 – 1) >2tan-13 = 2π3

B = 3sin-1(13) + sin-1(35)

   = sin-1[3. 13 - 4(13)3] + sin-1(35)

   = sin-1(2327) + sin-1(35)

   < sin-1(32) + sin-1(32) = 2π3 [∵2327 < 32 and 35 < 3 2]

So A > B.

20. Problem : Suppose a calculator is broken and the only keys that still work are the sin, cos, tan, sin−1, cos−1, and tan−1 buttons. The display initially shows 0. Given any positive rational number q, show that we can get q to appear on the display panel of the calculator by pressing some finite sequence of buttons. Assume that the calculator does real-number calculations with infinite precision, and that all functions are in terms of radians.
Solution: Because cos−1 sin θ = π2θ and tan(π2θ)= 1tanθ for 0 < θ < π2 , we have for any x > 0,

tan cos−1 sin tan−1 x =  tan(π2− tan−1x)= 1x.............................. (*)

Also, for x ≥ 0,

cos tan−1x = 1x+1 ,

so by (*),

tan cos−1 sin tan−1 cos tan−1x =x+1............................. (**)

By induction on the denominator of r, we now prove that √r, for every non-negative rational number r, can be obtained by using the operations

x x+1 and  x 1x.

If the denominator is 1, we can obtain √r, for every nonnegative integer r, by repeated application of √x →  x+1. Now assume that we can get√r for all rational numbers r with denominator up to n. In particular, we can get any of

n+11, n+12, ... n+1n,

so we can also get

1n+1, 2n+1, ... nn+1,

r, for any positive r of exact denominator n + 1, can be obtained by repeatedly applying √x  → x+1.
Thus for any positive rational number r, we can obtain √r. In particular, we
can obtain q2 = q.

Complete Tutorial (MCQ Quizzes after this):

MCQ Quiz #1

Companion MCQ Quiz #1 for Inverse Trigonometric Ratios (Challenging Problems) - test how much you know about the topic. Your score will be e-mailed to you at the address you provide.

MCQ Quiz- Inverse Trigonometric Ratio

MCQ Quiz #2

Companion MCQ Quiz #2 for Inverse Trigonometric Ratios (Challenging Problems) - test how much you know about the topic. Your score will be e-mailed to you at the address you provide.

MCQ Quiz: Inverse Trigonometric Functions- Challenging Problems

In case you'd like to take a look at other Trigonometry tutorials : 

Trigonometry 1a ( Introduction to Trigonometry - Definitions, Formulas ) Introducing trigonometric ratios, plots of trigonometric functions, compound angle formulas. Domains and ranges of trigonometric functions, monotonicity of trigonometric functions quadrant wise. Formulas for double and triple angle ratios.

Trigonometry 1b ( Tutorial with solved problems based on Trigonometric ratios ) Problems based on the concepts introduced above.
Trigonometry 2a ( Basic concepts related to Heights and Distances ) Applying trigonometry to problems involving heights and distances. Angles of elevation and depression. Sine and Cosine rule, half angle formulas. Circumradius, inradius and escribed radius. Circumcentre, incentre, centroid and median of a triangle.

Trigonometry 2b ( Tutorial with solved problems related to Heights and Distances and other applications of Trigonometry ) - Problems based on the concepts introduced above.

Trigonometry 3a ( Introducing Inverse Trigonometric Ratios)
Inverse trigonometric ratios - their domains, ranges and plots. 

Trigonometry 3b ( Tutorial with solved problems related to inverse trigonometric ratios )- Problems related to inverse trigonometric ratios. 

Trigonometry 4 ( A tutorial on solving trigonometric equations )- Solving trigonometric equations. Methods and transformations frequently used in solving such equations.