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Inverse Trigonometric Functions Solved PYQs & Practice Problems for NIMCET

Solve top previous year questions (PYQs) and practice problems on Inverse Trigonometric Functions from past NIMCET exams with complete step-by-step derivations.

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Updated 17 August 2026

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Inverse Trigonometric Functions: Solved PYQs & Practice Problems for NIMCET

Inverse Trigonometric Functions (ITF) are an important part of NIMCET Mathematics. Questions commonly test principal value ranges, inverse tangent addition formulas, telescoping series, domain restrictions, and complementary-angle identities.

This guide presents 6 high-yield solved problems with step-by-step solutions and shortcut insights.

Note: The mathematical solutions below have been verified. The exact claim that every question is an official NIMCET PYQ should be checked against the original NIMCET question paper before labeling it as a verified PYQ. The NIMCET syllabus explicitly includes principal values of inverse trigonometric functions under Trigonometry.

Solved Problem 1: Principal Value of an Inverse Trigonometric Function

Question:

The principal value of

cos⁡−1(cos⁡7π6)\cos^{-1}\left(\cos\frac{7\pi}{6}\right)

is:

  • (A) 7π6\frac{7\pi}{6}
  • (B) 5π6\frac{5\pi}{6}
  • (C) π6\frac{\pi}{6}
  • (D) −π6-\frac{\pi}{6}

Solution:

The principal value range of cos⁡−1x\cos^{-1}x is

[0,π].[0,\pi].

Since

7π6>π,\frac{7\pi}{6}>\pi,

the angle 7π6\frac{7\pi}{6} is outside the principal-value range.

Using

cos⁡(2π−θ)=cos⁡θ,\cos(2\pi-\theta)=\cos\theta,

we get

cos⁡7π6=cos⁡(2π−7π6)=cos⁡5π6.\cos\frac{7\pi}{6} = \cos\left(2\pi-\frac{7\pi}{6}\right) = \cos\frac{5\pi}{6}.

Since

5π6∈[0,π],\frac{5\pi}{6}\in[0,\pi],

we have

cos⁡−1(cos⁡7π6)=5π6.\cos^{-1}\left(\cos\frac{7\pi}{6}\right) = \frac{5\pi}{6}.

Correct Option: (B)

Shortcut

For cos⁡−1(cos⁡θ)\cos^{-1}(\cos\theta), always convert θ\theta into the principal range [0,π][0,\pi] before applying the inverse function.

Solved Problem 2: Sum of Inverse Tangents

Question:

If

tan⁡−12+tan⁡−13+θ=π,\tan^{-1}2+\tan^{-1}3+\theta=\pi,

then θ\theta is equal to:

  • (A) π4\frac{\pi}{4}
  • (B) π2\frac{\pi}{2}
  • (C) 3π4\frac{3\pi}{4}
  • (D) π3\frac{\pi}{3}

Solution:

For x>0x>0, y>0y>0, and xy>1xy>1,

tan⁡−1x+tan⁡−1y=π+tan⁡−1(x+y1−xy).\tan^{-1}x+\tan^{-1}y = \pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right).

Here,

x=2,y=3,x=2,\qquad y=3,

and

xy=6>1.xy=6>1.

Therefore,

tan⁡−12+tan⁡−13=π+tan⁡−1(2+31−6).\tan^{-1}2+\tan^{-1}3 = \pi+\tan^{-1}\left(\frac{2+3}{1-6}\right).

Hence,

=π+tan⁡−1(−1)= \pi+\tan^{-1}(-1) =π−π4=3π4.= \pi-\frac{\pi}{4} = \frac{3\pi}{4}.

Given

3π4+θ=π,\frac{3\pi}{4}+\theta=\pi,

we obtain

θ=π−3π4=π4.\theta = \pi-\frac{3\pi}{4} = \frac{\pi}{4}.

Correct Option: (A)

Shortcut

Both tan⁡−12\tan^{-1}2 and tan⁡−13\tan^{-1}3 are positive, and their sum is greater than π2\frac{\pi}{2}. Therefore, the answer cannot be −π4-\frac{\pi}{4}. The correct principal value is

3π4.\boxed{\frac{3\pi}{4}}.

Solved Problem 3: Domain of an Inverse Trigonometric Function

Question:

The domain of the function

f(x)=sin⁡−1(2x−3)f(x)=\sin^{-1}(2x-3)

is:

  • (A) [0,1][0,1]
  • (B) [1,2][1,2]
  • (C) [−1,1][-1,1]
  • (D) [1,3][1,3]

Solution:

For sin⁡−1u\sin^{-1}u to be defined for real values,

−1≤u≤1.-1\leq u\leq1.

Here,

u=2x−3.u=2x-3.

Therefore,

−1≤2x−3≤1.-1\leq2x-3\leq1.

Adding 33 throughout,

2≤2x≤4.2\leq2x\leq4.

Dividing by 22,

1≤x≤2.1\leq x\leq2.

Hence,

x∈[1,2].\boxed{x\in[1,2]}.

Correct Option: (B)

Shortcut

For

sin⁡−1(f(x))\sin^{-1}(f(x))

or

cos⁡−1(f(x)),\cos^{-1}(f(x)),

always impose

−1≤f(x)≤1.-1\leq f(x)\leq1.

Solved Problem 4: Complementary Inverse Trigonometric Functions

Question:

If

sin⁡−1x+sin⁡−1y=2π3,\sin^{-1}x+\sin^{-1}y=\frac{2\pi}{3},

then the value of

cos⁡−1x+cos⁡−1y\cos^{-1}x+\cos^{-1}y

is:

  • (A) π3\frac{\pi}{3}
  • (B) 2π3\frac{2\pi}{3}
  • (C) π6\frac{\pi}{6}
  • (D) π\pi

Solution:

For real x∈[−1,1]x\in[-1,1],

sin⁡−1x+cos⁡−1x=π2.\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}.

Similarly,

sin⁡−1y+cos⁡−1y=π2.\sin^{-1}y+\cos^{-1}y=\frac{\pi}{2}.

Adding the two equations,

sin⁡−1x+sin⁡−1y+cos⁡−1x+cos⁡−1y=π.\sin^{-1}x+\sin^{-1}y + \cos^{-1}x+\cos^{-1}y = \pi.

Given

sin⁡−1x+sin⁡−1y=2π3,\sin^{-1}x+\sin^{-1}y=\frac{2\pi}{3},

we get

2π3+cos⁡−1x+cos⁡−1y=π.\frac{2\pi}{3} + \cos^{-1}x+\cos^{-1}y = \pi.

Therefore,

cos⁡−1x+cos⁡−1y=π−2π3=π3.\cos^{-1}x+\cos^{-1}y = \pi-\frac{2\pi}{3} = \frac{\pi}{3}.

Correct Option: (A)

Shortcut

Whenever you see

sin⁡−1x+cos⁡−1x,\sin^{-1}x+\cos^{-1}x,

immediately replace it with

π2.\frac{\pi}{2}.

Solved Problem 5: Infinite Series of Inverse Tangents

Question:

The value of

∑n=1∞tan⁡−1(12n2)\sum_{n=1}^{\infty} \tan^{-1}\left(\frac{1}{2n^2}\right)

is:

  • (A) π4\frac{\pi}{4}
  • (B) π2\frac{\pi}{2}
  • (C) π3\frac{\pi}{3}
  • (D) π6\frac{\pi}{6}

Solution:

Consider the identity

tan⁡−1a−tan⁡−1b=tan⁡−1(a−b1+ab),\tan^{-1}a-\tan^{-1}b = \tan^{-1}\left(\frac{a-b}{1+ab}\right),

provided the principal-value condition is satisfied.

Take

a=2n+1,b=2n−1.a=2n+1,\qquad b=2n-1.

Then

a−b=2a-b=2

and

1+ab=1+(2n+1)(2n−1)=1+(4n2−1)=4n2.1+ab = 1+(2n+1)(2n-1) = 1+(4n^2-1) = 4n^2.

Therefore,

tan⁡−1(2n+1)−tan⁡−1(2n−1)=tan⁡−1(24n2)\tan^{-1}(2n+1)-\tan^{-1}(2n-1) = \tan^{-1}\left(\frac{2}{4n^2}\right) =tan⁡−1(12n2).= \tan^{-1}\left(\frac{1}{2n^2}\right).

Hence,

∑n=1Ntan⁡−1(12n2)=∑n=1N[tan⁡−1(2n+1)−tan⁡−1(2n−1)].\sum_{n=1}^{N} \tan^{-1}\left(\frac{1}{2n^2}\right) = \sum_{n=1}^{N} \left[ \tan^{-1}(2n+1)-\tan^{-1}(2n-1) \right].

Expanding,

SN=(tan⁡−13−tan⁡−11)+(tan⁡−15−tan⁡−13)+⋯+(tan⁡−1(2N+1)−tan⁡−1(2N−1)).S_N = (\tan^{-1}3-\tan^{-1}1) + (\tan^{-1}5-\tan^{-1}3) +\cdots + (\tan^{-1}(2N+1)-\tan^{-1}(2N-1)).

All intermediate terms cancel:

SN=tan⁡−1(2N+1)−tan⁡−11.S_N = \tan^{-1}(2N+1)-\tan^{-1}1.

As N→∞N\to\infty,

tan⁡−1(2N+1)→π2.\tan^{-1}(2N+1)\to\frac{\pi}{2}.

Also,

tan⁡−11=π4.\tan^{-1}1=\frac{\pi}{4}.

Therefore,

S∞=π2−π4=π4.S_\infty = \frac{\pi}{2}-\frac{\pi}{4} = \frac{\pi}{4}.

Correct Option: (A)

Shortcut

Look for expressions of the form

tan⁡−1(A)−tan⁡−1(B)\tan^{-1}(A)-\tan^{-1}(B)

when an infinite inverse-tangent series contains rational terms. Such series often telescope.

Solved Problem 6: Double-Angle Transformation

Question:

If

x=15,x=\frac{1}{5},

then the value of

cos⁡(2cos⁡−1x+sin⁡−1x)\cos\left(2\cos^{-1}x+\sin^{-1}x\right)

is:

  • (A) −2425-\sqrt{\frac{24}{25}}
  • (B) −15-\frac{1}{5}
  • (C) 15\frac{1}{5}
  • (D) 00

Solution:

Using

sin⁡−1x+cos⁡−1x=π2,\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2},

we can write

2cos⁡−1x+sin⁡−1x=cos⁡−1x+(cos⁡−1x+sin⁡−1x).2\cos^{-1}x+\sin^{-1}x = \cos^{-1}x+ \left(\cos^{-1}x+\sin^{-1}x\right).

Therefore,

2cos⁡−1x+sin⁡−1x=cos⁡−1x+π2.2\cos^{-1}x+\sin^{-1}x = \cos^{-1}x+\frac{\pi}{2}.

Hence,

cos⁡(2cos⁡−1x+sin⁡−1x)=cos⁡(π2+cos⁡−1x).\cos\left(2\cos^{-1}x+\sin^{-1}x\right) = \cos\left(\frac{\pi}{2}+\cos^{-1}x\right).

Using

cos⁡(π2+θ)=−sin⁡θ,\cos\left(\frac{\pi}{2}+\theta\right)=-\sin\theta,

we get

=−sin⁡(cos⁡−1x).=-\sin(\cos^{-1}x).

Let

θ=cos⁡−1x.\theta=\cos^{-1}x.

Then

cos⁡θ=x=15.\cos\theta=x=\frac{1}{5}.

Since θ∈[0,π]\theta\in[0,\pi],

sin⁡θ=1−cos⁡2θ.\sin\theta = \sqrt{1-\cos^2\theta}.

Thus,

sin⁡θ=1−125=2425.\sin\theta = \sqrt{1-\frac{1}{25}} = \sqrt{\frac{24}{25}}.

Therefore,

−2425.\boxed{-\sqrt{\frac{24}{25}}}.

Correct Option: (A)

Frequently Asked Questions (FAQ)

Q1: Why is tan⁡−12+tan⁡−13\tan^{-1}2+\tan^{-1}3 equal to 3π4\frac{3\pi}{4} and not −π4-\frac{\pi}{4}?

Both 22 and 33 are positive, so

tan⁡−12>0andtan⁡−13>0.\tan^{-1}2>0 \quad\text{and}\quad \tan^{-1}3>0.

Moreover,

tan⁡−12>π4andtan⁡−13>π4.\tan^{-1}2>\frac{\pi}{4} \quad\text{and}\quad \tan^{-1}3>\frac{\pi}{4}.

Therefore, their sum is greater than π2\frac{\pi}{2}.

Using

tan⁡−1x+tan⁡−1y=π+tan⁡−1(x+y1−xy)\tan^{-1}x+\tan^{-1}y = \pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)

for x,y>0x,y>0 and xy>1xy>1,

tan⁡−12+tan⁡−13=π+tan⁡−1(−1)=3π4.\tan^{-1}2+\tan^{-1}3 = \pi+\tan^{-1}(-1) = \frac{3\pi}{4}.

The value −π4-\frac{\pi}{4} is only the value returned by the basic tangent-addition expression before correcting for the appropriate quadrant.

Q2: What is the domain of

f(x)=cos⁡−1(x2−4)?f(x)=\cos^{-1}(x^2-4)?

For cos⁡−1u\cos^{-1}u to be defined,

−1≤u≤1.-1\leq u\leq1.

Therefore,

−1≤x2−4≤1.-1\leq x^2-4\leq1.

Adding 44,

3≤x2≤5.3\leq x^2\leq5.

Hence,

x∈[−5,−3]∪[3,5].\boxed{ x\in[-\sqrt5,-\sqrt3]\cup[\sqrt3,\sqrt5] }.

Q3: How many ITF questions appear in NIMCET each year?

There is no fixed number of inverse-trigonometric-function questions prescribed by the NIMCET syllabus.

The official syllabus includes principal values of inverse trigonometric functions under Trigonometry, but the exact number of questions from ITF can vary from year to year.

Therefore, it is better to prepare ITF as part of the broader Trigonometry section rather than assuming a fixed yearly question count.

Q4: Can I use

sin⁡−1x+cos⁡−1x=π2\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}

for complex numbers?

No.

This identity is used for real values of xx in the domain

−1≤x≤1.-1\leq x\leq1.

For NIMCET preparation, inverse-trigonometric identities should be applied within their real-domain and principal-value restrictions.

Important ITF Formulas for NIMCET

Principal Value Ranges

sin⁡−1x∈[−π2,π2]\sin^{-1}x\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right] cos⁡−1x∈[0,π]\cos^{-1}x\in[0,\pi] tan⁡−1x∈(−π2,π2)\tan^{-1}x\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)

Complementary Identity

sin⁡−1x+cos⁡−1x=π2,−1≤x≤1\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}, \qquad -1\leq x\leq1

Inverse Tangent Addition

For x,y>0x,y>0 and xy>1xy>1,

tan⁡−1x+tan⁡−1y=π+tan⁡−1(x+y1−xy).\tan^{-1}x+\tan^{-1}y = \pi+ \tan^{-1}\left(\frac{x+y}{1-xy}\right).

For xy<1xy<1,

tan⁡−1x+tan⁡−1y=tan⁡−1(x+y1−xy)\tan^{-1}x+\tan^{-1}y = \tan^{-1}\left(\frac{x+y}{1-xy}\right)

subject to the appropriate principal-value conditions.

Domain Rules

For

sin⁡−1(f(x))\sin^{-1}(f(x))

and

cos⁡−1(f(x)),\cos^{-1}(f(x)),

we require

−1≤f(x)≤1.-1\leq f(x)\leq1.

For

tan⁡−1(f(x)),\tan^{-1}(f(x)),

any real value of f(x)f(x) is allowed, provided f(x)f(x) itself is defined.

Key Takeaways for NIMCET

  1. Always remember the principal-value ranges.
  2. Do not blindly apply inverse-tangent addition formulas without checking the quadrant.
  3. For sin⁡−1(f(x))\sin^{-1}(f(x)) and cos⁡−1(f(x))\cos^{-1}(f(x)), immediately impose −1≤f(x)≤1.-1\leq f(x)\leq1.
  4. Look for telescoping patterns in inverse-tangent series.
  5. Use sin⁡−1x+cos⁡−1x=π2\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2} whenever applicable.
  6. In NIMCET, speed matters—recognizing these standard transformations can save significant calculation time.