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Inverse trigonometrical functions and their properties Short Tricks for JEE

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Short tricks for Inverse trigonometrical functions and their properties work only with strong fundamentals. Apply the tips below in timed sets and review every explanation.

Use these Inverse trigonometrical functions and their properties shortcuts to save time in JEE Mathematics papers — then validate speed with 14+ MCQs on Goodmarks.

Short tricks for speed

  • Inverse trigonometrical functions and their properties focus drill

    Solve 15 mixed MCQs for Inverse trigonometrical functions and their properties, review every explanation, and note formulas you hesitated on.

  • Calculus substitution scan

    Spot standard forms (sin²x, 1/(a²+x²), e^ax) before integrating — JEE rewards pattern recognition.

  • Graph sketch shortcut

    For coordinate geometry, mark intercepts and asymptotes first; many MCQs need only qualitative graph features.

Syllabus context

Part of Trigonometry in JEE Main Mathematics.

Free sample questions

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Q1MathsUnit 14: Trigonometry
Assertion Consider f(x)=sin⁡−1(sec⁡(tan⁡−1x)+\boldsymbol{f}(\boldsymbol{x})=\sin ^{-1}\left(\sec \left(\tan ^{-1} \boldsymbol{x}\right)+\right. cos⁡−1(cosec⁡(cot⁡−1x)\cos ^{-1}\left(\operatorname{cosec}\left(\cot ^{-1} x\right)\right. Statement-1: Domain of f(x)f(x) is a singleton. Reason Statement-2: Range of the function f(x)\boldsymbol{f}(\boldsymbol{x}) is a singleton.
Q2MathsUnit 14: Trigonometry
The number of real solutions of the equation tan⁡−1x(x+1)+\tan ^{-1} \sqrt{x(x+1)}+ sin⁡−1x2+x+1=π2\sin ^{-1} \sqrt{x^{2}+x+1}=\frac{\pi}{2} is
Q3MathsUnit 14: Trigonometry
If value of x\mathbf{x} which satisfy equation (cot⁡−1x)2−3(cot⁡−1x)+2>0\left(\cot ^{-1} x\right)^{2}-3\left(\cot ^{-1} x\right)+2>0 is x<x< cot⁡a\cot a or x>cot⁡bx>\cot b Find the value of a+ba+b
Q4MathsUnit 14: Trigonometry
Statement I: The equation (sin⁡−1x)3+(cos⁡−1x)3−aπ3=0\left(\sin ^{-1} x\right)^{3}+\left(\cos ^{-1} x\right)^{3}-a \pi^{3}=0 has solution for all a⩾132a \geqslant \frac{1}{32} Statement II : For any xϵR,sin−1x+\boldsymbol{x} \boldsymbol{\epsilon} \boldsymbol{R}, \boldsymbol{s} \boldsymbol{i n}^{-1} \boldsymbol{x}+ cos⁡−1x=π2\cos ^{-1} x=\frac{\pi}{2} and 0≤(sin⁡−1x−π4)2≤0 \leq\left(\sin ^{-1} x-\frac{\pi}{4}\right)^{2} \leq 9π216\frac{9 \pi^{2}}{16}
Q5MathsUnit 14: Trigonometry
Assertion (A)(A) If 0<x<π20<x<\frac{\pi}{2} then sin⁡−1(cos⁡x)+cos⁡−1(sin⁡x)=π−2x\sin ^{-1}(\cos x)+\cos ^{-1}(\sin x)=\pi-2 x Reason (R)cos⁡−1x=π2−sin⁡−1x∀x∈(\mathrm{R}) \cos ^{-1} x=\frac{\pi}{2}-\sin ^{-1} x \forall x \in [0,1][\mathbf{0}, \mathbf{1}]
Q6MathsUnit 14: Trigonometry
Assertion fi=12nsin⁡−1xi=nπ∀nϵNf_{i=1}^{2 n} \sin ^{-1} x_{i}=n \pi \forall n \epsilon N then ∑i=12nxi=\sum_{i=1}^{2 n} x_{i}= ∑i=12nxi2=∑i=12nxin=2n\sum_{i=1}^{2 n} x_{i}^{2}=\sum_{i=1}^{2 n} x_{i}^{n}=2 n Reason −π2≤sin⁡−1x≤π2∀xϵ[−1,1]-\frac{\pi}{2} \leq \sin ^{-1} x \leq \frac{\pi}{2} \forall x \epsilon[-1,1]
Q7MathsUnit 14: Trigonometry
Find the value of sin⁡−1x+sin⁡−11x+\sin ^{-1} x+\sin ^{-1} \frac{1}{x}+ cos⁡−1x+cos⁡−11x\cos ^{-1} x+\cos ^{-1} \frac{1}{x}
Q8MathsUnit 14: Trigonometry
If [sin⁡−1cos⁡−1sin⁡−1tan⁡−1θ]=1,\left[\sin ^{-1} \cos ^{-1} \sin ^{-1} \tan ^{-1} \theta\right]=1, where [.] denotes the greatest integer function, the θ\theta lies in the interval

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No — tricks complement concepts. Master the theory first, then use shortcuts in timed MCQ practice.

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