46 JEE Main practice questions on Trigonometry (Identities, Equations, Inverse Functions, Solution of Triangles), part of Mathematics. Below are 12 of them in full, each with the answer and a written explanation.
Answer: x < 1
Using the subtraction formula tan⁻¹ u - tan⁻¹ v = tan⁻¹((u-v)/(1+uv)) requires uv > -1. Here u = (1+x)/(1-x), v = x, so uv = x(1+x)/(1-x). For x < 1, 1-x > 0 and the condition uv > -1 simplifies to x² > -1, always true. For x > 1, the principal value of tan⁻¹((1+x)/(1-x)) shifts by π, so the identity becomes π/4 - π. Hence the identity holds exactly for x < 1.
Answer: 1
Take sine both sides: 2x = sin(π/3)cos(sin⁻¹ x) - cos(π/3)x = (√3/2)√(1-x²) - x/2. Rearranging gives 5x/2 = (√3/2)√(1-x²); squaring yields 25x²/4 = (3/4)(1-x²) → 28x²/4 = 3/4 → x² = 3/28 → x = ±√(3/28). Check x = -√(3/28): sin⁻¹ x and sin⁻¹(2x) are both negative, sum cannot be π/3. So only x = √(3/28) = √21/14 is valid. Hence exactly one solution.
Answer: It has horizontal asymptotes y = ±π/2.
The graph of tan⁻¹ x is the reflection of tan x restricted to (-π/2, π/2). As x → ∞, tan⁻¹ x → π/2; as x → -∞, tan⁻¹ x → -π/2, giving horizontal asymptotes y = ±π/2. The function is strictly increasing, not decreasing. Its domain is all real numbers, not just |x| ≤ 1. It passes through (0,0), not (0, π/4).
Answer: π
Using the two-term formula, tan⁻¹ 2 + tan⁻¹ 3 = π + tan⁻¹((2+3)/(1-6)) = π + tan⁻¹(-1) = 3π/4. Adding tan⁻¹ 1 = π/4 gives π. Alternatively, the three-term formula yields numerator 0 and denominator -10, so tan⁻¹(0) = 0, but since denominator < 0 and arguments positive, add π to get π.
Answer: x ∈ [-1, 1]
The identity 2 tan⁻¹ x = sin⁻¹(2x/(1+x²)) holds when the output of 2 tan⁻¹ x lies in the principal range of sin⁻¹, i.e., [-π/2, π/2]. Since tan⁻¹ x ∈ (-π/2, π/2), we need |2 tan⁻¹ x| ≤ π/2, which gives |tan⁻¹ x| ≤ π/4, i.e., |x| ≤ 1. Thus the domain is x ∈ [-1, 1].
Answer: sin⁻¹(-x) = - sin⁻¹ x for all x ∈ [-1,1]
sin⁻¹(-x) = - sin⁻¹ x holds for x ∈ [-1,1] because sin is odd on its principal branch [-π/2, π/2]. For example, sin⁻¹(-1/2) = -π/6 = - sin⁻¹(1/2). The other options are false: cos⁻¹(-x) = π - cos⁻¹ x, tan⁻¹(-x) = - tan⁻¹ x, and cosec⁻¹(-x) = - cosec⁻¹ x.
Answer: 0
Let u = (x-1)/(x-2), v = (x+1)/(x+2). Using tan^-1 u + tan^-1 v = tan^-1((u+v)/(1-uv)) with uv < 1, we get (u+v)/(1-uv) = 1. Simplifying yields -2(x^2-2)/3 = 1, so x^2 = 1/2, giving x = ±1/√2. Both satisfy uv < 1. Sum of solutions = 0.
Answer: 0
Substitute x = -1/2: 4(-1/2)³ - 3(-1/2) = 4(-1/8) + 3/2 = -1/2 + 3/2 = 1. Then cos⁻¹(1) = 0. The identity cos⁻¹(4x³-3x) = 3 cos⁻¹ x holds only for x ∈ [1/2, 1]; here x = -1/2 is outside that interval, so direct substitution is valid.
Answer: π
Let a=1, b=2, c=3. Using tan⁻¹ a + tan⁻¹ b + tan⁻¹ c = tan⁻¹((a+b+c-abc)/(1-ab-bc-ca)) + kπ. Numerator = 1+2+3-6=0, denominator = 1-2-6-3=-10. Since denominator < 0 and all numbers positive, we add π, giving tan⁻¹(0)+π = π.
Answer: [π^2/8, 5π^2/4]
Using sin⁻¹x + cos⁻¹x = π/2, let t = sin⁻¹x ∈ [-π/2, π/2]. Then f = t² + (π/2 - t)² = 2t² - πt + π²/4. Vertex at t = π/4 gives minimum π²/8. Endpoints: t = -π/2 gives 5π²/4, t = π/2 gives π²/4. So range is [π²/8, 5π²/4].
Answer: 1/2
Using cos^-1 x = π/2 - sin^-1 x, equation becomes 2 sin^-1 x + sin^-1(1-x) = π/2. Testing x=0 gives LHS=0+π/2=π/2, works. Testing x=1/2 gives π/6+π/6=π/3, RHS=π/3, works. Both satisfy domain. Sum = 0 + 1/2 = 1/2.
Answer: 1
Using formula, (2x+3x)/(1-6x^2) = 1 gives 6x^2+5x-1=0, roots x=1/6 and x=-1. Check xy=6x^2: for x=1/6, xy=1/6<1, valid; for x=-1, xy=6≥1, so formula gives π+tan^-1...) not π/4, thus rejected. Only one solution.
This page shows 12 of 46 questions on this topic. The full set, with progress tracking and five agent perspectives per question, is in the JupiteX app — browse the exam catalogue or browse the Learn library.