Inverse Trigonometric Functions Questions with Solutions

21 questions4 easy13 medium4 hard

Inverse trigonometric functions, such as sin^-1 x, cos^-1 x and tan^-1 x, return the angle whose trigonometric ratio is a given value. Because trigonometric functions repeat, each inverse is defined only on a principal value branch. For example, sin^-1 x takes values between -pi/2 and pi/2. This topic covers domains and principal value ranges, finding principal values, and key properties such as sin^-1 x + cos^-1 x = pi/2 and the formulas for tan^-1 x + tan^-1 y. Exam questions test simplifying expressions, solving equations, and evaluating compositions such as sin(cos^-1 x). The most common mistake is ignoring the principal range, for example assuming sin^-1(sin x) always equals x. Practise the inverse trigonometric functions questions below, each with a step-by-step solution.

All Questions Easy Medium Hard

Q1.
The number of solutions of the equation $t an^{- 1} (3 x) + t an^{- 1} (2 x) = \frac{π}{4}$ is:
A. 0B. 2C. 1D. Infinitely many
mediumAsked in NIMCET 2026View solution ›
Q2.
The valid range of $x$ for which $cos^{- 1} (2 x - 3)$ is defined is
A. $[1, 2]$ B. $[\frac{1}{5}, 2]$ C. $[\frac{1}{5}, 1]$ D. $[- 1, 1]$
mediumAsked in MAH MCA CET 2025View solution ›
Q3.
If $t an^{- 1} (\frac{1 - x}{1 + x}) = \frac{1}{2} t an^{- 1} x$ , then $x$ is
A. $\frac{1}{2}$ B. $\frac{1}{3}$ C. $3$ D. 2
mediumAsked in MAH MCA CET 2025View solution ›
Q4.
If $t an^{- 1} (\frac{1 - x}{1 + x}) = \frac{1}{2} t an^{- 1} x$ , then $x$ is
A. $\frac{1}{2}$ B. $\frac{1}{3}$ C. $3$ D. 2
mediumAsked in MAH MCA CET 2025View solution ›
Q5.
If $n_{1}$ and $n_{2}$ are the numbers of real solutions of $x = s in^{- 1} x$ and $x = s in x$ respectively, then the value of $n_{2} - n_{1}$ is
A. 1B. 0C. 2D. 3
mediumAsked in NIMCET 2023View solution ›
Q6.
The value of $co t^{- 1} (21) + co t^{- 1} (13) + co t^{- 1} (- 8)$ is
A. 0B. $π$ C. $\infty$ D. $\frac{π}{2}$
mediumAsked in NIMCET 2012View solution ›
Q7.
If $s in^{- 1} \frac{2 a}{1 + a ^{2}} - cos^{- 1} \frac{1 - b ^{2}}{1 + b ^{2}} = t an^{- 1} \frac{2 x}{1 - x ^{2}}$ , then $x$ is equal to
A. $a$ B. $b$ C. $\frac{a + b}{1 - ab}$ D. $\frac{a - b}{1 + ab}$
mediumAsked in NIMCET 2010View solution ›
Q8.
If $θ = t an^{- 1} \frac{1}{1 + 1 \cdot 2} + t an^{- 1} \frac{1}{1 + 2 \cdot 3} + t an^{- 1} \frac{1}{1 + 3 \cdot 4} + \dots + t an^{- 1} \frac{1}{1 + n ( n + 1 )}$ , then $t an θ$ is equal to:
A. $\frac{n}{n + 1}$ B. $\frac{n + 1}{n + 2}$ C. $\frac{n}{n + 2}$ D. $\frac{n - 1}{n + 2}$
mediumAsked in NIMCET 2009View solution ›
Q9.
If $t an^{- 1} (2 x) + t an^{- 1} (3 x) = \frac{π}{4}$ , then $x$ is:
A. $\frac{1}{6}$ B. $\frac{1}{3}$ C. $\frac{1}{2}$ D. $\frac{1}{4}$
mediumAsked in NIMCET 2009View solution ›
Q10.
The number of solutions of $t an^{- 1} x (x + 1) + s in^{- 1} x^{2} + x + 1 = \frac{π}{2}$ is:
A. zeroB. oneC. twoD. infinite
mediumAsked in NIMCET 2009View solution ›
Q11.
If $y = sec^{- 1} (\frac{x + 1}{x - 1}) + s in^{- 1} (\frac{x - 1}{x + 1})$ , $x \in [0, \infty)$ and $x \neq = 1$ , then $\frac{d y}{d x}$ is equal to
A. $1$ B. $\frac{x - 1}{x + 1}$ C. $0$ D. $\frac{x + 1}{x - 1}$
mediumAsked in NIMCET 2008View solution ›
Q12.
The value of $2 t an^{- 1} [cosec (t an^{- 1} x) - t an (co t^{- 1} x)]$ is
A. $t an x$ B. $co t x$ C. $t an^{- 1} x$ D. $cosec^{- 1} x$
mediumAsked in NIMCET 2020View solution ›
Q13.
The value of $co t (cosec^{- 1} \frac{5}{3} + t an^{- 1} \frac{2}{3})$ is
A. $\frac{6}{17}$ B. $\frac{3}{17}$ C. $\frac{4}{17}$ D. $\frac{5}{17}$
mediumAsked in NIMCET 2022 +1View solution ›