Inverse Trigonometric Functions
9 Board Maths previous year questions on Inverse Trigonometric Functions — options free on every question; 1 include the answer & explanation free, the rest unlock with PYQ Pass.
The number of solutions of \({\text{tan}}^{−1}4x+{\text{tan}}^{−1}6x=\frac{\pi }{6}\), where \(−\frac{1}{2\sqrt{6}} [JEE Main 2026, 22 Jan (Shift 1)]
1
\({\text{tan}}^{−1}4x+{\text{tan}}^{−1}6x=\frac{\pi }{6}\)
\(\Rightarrow {\text{tan}}^{−1}\left(\frac{4x+6x}{1-24{x}^{2}}\right)=\frac{\pi }{6}\)
\(\Rightarrow \frac{10x}{1−24{x}^{2}}=\frac{1}{\sqrt{3}}\)
\(\Rightarrow 24{x}^{2}+10\sqrt{3}x−1=0\)
\(x=\frac{−10\sqrt{3}\pm \sqrt{300+96}}{48}\)
\(x=\frac{\sqrt{396}−10\sqrt{3}}{48}\)
Only 1 solution in \(\left(−\frac{1}{2\sqrt{6}},\frac{1}{2\sqrt{6}}\right)\)
If \(f(x)=16\left(\left(\sec ^{-1} x\right)^2+\left(\operatorname{cosec}^{-1} x\right)^2\right)\) then the sum of max. and min. value of \(f(x)\) is (22 Jan, Shift I, Memory Based)
Options are free to see. Unlock the correct answer and full explanation with Pass.
If \(\mathrm{y}={\sin }^{-1}x,-1\leq x\leq 0\), then the range of y is:
Options are free to see. Unlock the correct answer and full explanation with Pass.
If \(\mathrm{y}={\sin }^{-1}x,-1\leq x\leq 0\), then the range of \(y\) is:
Options are free to see. Unlock the correct answer and full explanation with Pass.
(ii) The principal value of \({\sin }^{-1}\left(-\frac{1}{2}\right)\) is -
Options are free to see. Unlock the correct answer and full explanation with Pass.
The principal value of \({\sin }^{-1}\left(\frac{1}{\sqrt{2}}\right)\)is
Options are free to see. Unlock the correct answer and full explanation with Pass.
Assertion (A) : Domain of \(\mathrm{y}={\cos }^{-1}(\mathrm{x})\) is \([-1,1]\).
Reason (R) : The range of the principal value branch of \(\mathrm{y}={\cos }^{-1}(x)\) is \([0,\pi ]-\left\{\frac{\pi }{2}\right\}\).
Options are free to see. Unlock the correct answer and full explanation with Pass.
Direction : Two statements are given one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer from the following options :
Assertion (A) : Domain of \(\mathrm{y}={\cos }^{-1}(\mathrm{x})\) is \([-1,1]\).
Reason (R) : The range of the principal value branch of \(\mathrm{y}={\cos }^{-1}(x)\) is \([0,\pi ]-\left\{\frac{\pi }{2}\right\}\).
Options are free to see. Unlock the correct answer and full explanation with Pass.
\({\tan }^{-1}\sqrt{3}-{\sec }^{-1}(-2)\) is equal to
Options are free to see. Unlock the correct answer and full explanation with Pass.
Practice more Board Maths PYQs
Browse every Maths chapter, or explore the full Board question bank.
All Maths chapters →