Question types

Inverse Trigonometric Functions question types

262 questions across 4 question groups — pick any mix to generate a Maths paper with step-by-step answer keys.

262
Questions
4
Question groups
5
Question types
Sample Questions

Inverse Trigonometric Functions questions

One sample from each question group in this chapter. Select any group above to see the full set with answer keys.

Q 1MCQ1 Mark
$\sin\Big\{2\cos^{-1}\Big(\frac{-3}{5}\Big)\Big\}$ is equal to:
  • A
    $\frac{6}{25}$
  • B
    $\frac{24}{25}$
  • C
    $\frac{4}{5}$
  • $-\frac{24}{25}$

Answer: D.

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Q 2MCQ1 Mark
If $+\cos^{-1}\text{x}>\sin^{-1},$ then:
  • $\frac{1}{\sqrt2}<\text{x}\leq1$
  • B
    $0\leq\text{x}\leq\frac{1}{\sqrt2}$
  • C
    $-1\leq\text{x}<\frac{1}{\sqrt2}$
  • D
    $\text{x}>0$

Answer: A.

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Q 3MCQ1 Mark
If $\tan^{-1}\frac{\text{x}+1}{\text{x}-1}+\tan^{-1}\frac{\text{x}-1}{\text{x}}=\tan^{-1}(-7),$ then the value of $x$ is:
  • A
    $0$
  • B
    $-2$
  • C
    $1$
  • $2$

Answer: D.

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Q 4MCQ1 Mark
The number of solutions of the equation $\tan^{-1}2\text{x}+\tan^{-1}3\text{x}=\frac{\pi}{4}$ is:
  • A
    $2$
  • B
    $3$
  • C
    $1$
  • none of these

Answer: D.

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Q 5MCQ1 Mark
If $\sin^{-1}\Big(\frac{2\text{a}}{1-\text{a}^2}\Big)+\cos^{-1}\Big(\frac{1-\text{a}^2}{1+\text{a}^2}\Big)=\tan^{-1}\Big(\frac{2\text{x}}{1-\text{x}^2}\Big),$ where $\text{a},\text{x}\in(0,1),$ then the value of $x$ is:
  • A
    $0$
  • B
    $\frac{\text{a}}{2}$
  • C
    $\text{a}$
  • $\frac{2\text{a}}{1-\text{a}^2}$

Answer: D.

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For any a, b, x, y > 0, prove that:
$\frac{2}{3}\tan^{-1}\Big(\frac{3\text{a}\text{b}^2-\text{a}^3}{\text{b}^3-3\text{a}^2\text{b}}\Big)+\frac{2}{3}\tan^{-1}\Big(\frac{3\text{x}\text{y}^2-\text{x}^3}{\text{y}^3-3\text{x}^2\text{y}}\Big)=\tan^{-1}\frac{2\alpha\beta}{\alpha^2-\beta^2}$
where $\alpha=-\text{ax}+\text{by},\beta=\text{bx}+\text{ay}$
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Prove that: $\tan^{-1}\frac{2\text{a}\text{b}}{\text{a}^2-\text{b}^2}+\tan^{-1}\frac{2\text{xy}}{\text{x}^2-\text{y}^2}=\tan^{-1}\frac{2\alpha\beta}{\alpha^2-\beta^2},$where $\alpha=\text{ax}-\text{by}$ and $\beta=\text{ay}+\text{bx}.$
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