Questions are transcribed from NTA's official question paper. Diagrams are described in words; open the official PDF to see them. Every answer was solved twice independently and is shown only when both agree. Where they did not, we say so.
Let $0 < \alpha < 1$, $\beta = \frac{1}{3\alpha}$ and $\tan^{-1}(1-\alpha) + \tan^{-1}(1-\beta) = \frac{\pi}{4}$. Then $6(\alpha + \beta)$ is equal to:
(1) 6
(2) 7
(3) 8
(4) 9
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Answer: Option (2)
Key idea: Inverse Trigonometric Functions - Addition Formula
Step 1: Use the formula $\tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1}\left(\frac{a+b}{1-ab}\right)$ with $a = 1-\alpha$ and $b = 1-\beta$
Step 2: Since the sum equals $\frac{\pi}{4}$, we have $\frac{(1-\alpha)+(1-\beta)}{1-(1-\alpha)(1-\beta)} = 1$
Step 3: Simplify to get $\frac{2-\alpha-\beta}{\alpha+\beta-\alpha\beta} = 1$, which gives $2 = 2(\alpha+\beta) - \alpha\beta$
If $\sin(\tan^{-1}(x\sqrt{2})) = \cot(\sin^{-1}\sqrt{1-x^2})$, $x \in (0, 1)$, then the value of $x$ is:
(1) $\frac{1}{2}$
(2) $\frac{1}{3}$
(3) $\frac{2}{3}$
(4) $\frac{5}{8}$
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Answer: Option (1)
Key idea: Inverse Trigonometric Functions - Simplification and Equation Solving
Let $\theta = \tan^{-1}(x\sqrt{2})$, so $\tan\theta = x\sqrt{2}$. Using a right triangle, $\sin(\tan^{-1}(x\sqrt{2})) = \frac{x\sqrt{2}}{\sqrt{2x^2+1}}$
Let $\phi = \sin^{-1}\sqrt{1-x^2}$, so $\sin\phi = \sqrt{1-x^2}$ and $\cos\phi = x$. Thus $\cot(\sin^{-1}\sqrt{1-x^2}) = \frac{x}{\sqrt{1-x^2}}$
Equate both sides: $\frac{x\sqrt{2}}{\sqrt{2x^2+1}} = \frac{x}{\sqrt{1-x^2}}$. Divide by $x$ (since $x > 0$): $\frac{\sqrt{2}}{\sqrt{2x^2+1}} = \frac{1}{\sqrt{1-x^2}}$
Let $\alpha = 3 \sin^{-1}\left(\frac{6}{11}\right)$ and $\beta = 3 \cos^{-1}\left(\frac{4}{9}\right)$, where inverse trigonometric functions take only the principal values.
Given below are two statements:
**Statement I:** $\cos(\alpha + \beta) > 0$.
**Statement II:** $\cos(\alpha) < 0$.
In the light of the above statements, choose the correct answer from the options given below:
(1) Both Statement I and Statement II are true
(2) Both Statement I and Statement II are false
(3) Statement I is true but Statement II is false
Show answer and solution
Answer: Option (1)
Key idea: Range analysis of inverse trigonometric functions and their multiples
Find range of $\alpha = 3\sin^{-1}(6/11)$: Since $\sin^{-1}(6/11) \approx 0.576$ rad, we get $\alpha \approx 1.728$ rad
Since $\pi/2 \approx 1.571 < \alpha < \pi \approx 3.142$, we have $\alpha \in (\pi/2, \pi)$, so $\cos(\alpha) < 0$ (Statement II is TRUE)
Find range of $\beta = 3\cos^{-1}(4/9)$: Since $\cos^{-1}(4/9) \approx 1.107$ rad, we get $\beta \approx 3.321$ rad
Since $\pi \approx 3.142 < \beta < 3\pi/2 \approx 4.712$, we have $\beta \in (\pi, 3\pi/2)$
Calculate $\alpha + \beta \approx 1.728 + 3.321 = 5.049$ rad, which lies in $(3\pi/2, 2\pi)$
Since $\alpha + \beta \in (3\pi/2, 2\pi)$, we have $\cos(\alpha + \beta) > 0$ (Statement I is TRUE)
Both statements are true, so the answer is option (1)
FAQs
Frequently asked questions
How many questions come from Inverse Trigonometric Functions in JEE Main?
In the 9 shifts of JEE Main 2026 Session 2 we found 4 questions on it, about 0.4 per shift.
Where can I practise Inverse Trigonometric Functions PYQs for free?
On this page, with solutions. For more, use NTA Abhyas and our free mock tests.
