April 2026

Value of sin⁻¹(1/3) − cos⁻¹(−1/3) Question Evaluate: \[ \sin^{-1}\left(\frac{1}{3}\right) – \cos^{-1}\left(-\frac{1}{3}\right) \] Solution We use identity: \[ \cos^{-1}(-x) = \pi – \cos^{-1}(x) \] So, \[ \cos^{-1}\left(-\frac{1}{3}\right) = \pi – \cos^{-1}\left(\frac{1}{3}\right) \] Now substitute: \[ \sin^{-1}\left(\frac{1}{3}\right) – \left(\pi – \cos^{-1}\left(\frac{1}{3}\right)\right) \] \[ = \sin^{-1}\left(\frac{1}{3}\right) + \cos^{-1}\left(\frac{1}{3}\right) – \pi \] Using identity: \[ \sin^{-1}x + \cos^{-1}x = […]

If sin⁻¹(1/3) + cos⁻¹x = π/2, find x Question If \[ \sin^{-1}\left(\frac{1}{3}\right) + \cos^{-1}x = \frac{\pi}{2} \] Find \( x \). Solution We use identity: \[ \sin^{-1}a + \cos^{-1}a = \frac{\pi}{2} \] Comparing, \[ \cos^{-1}x = \frac{\pi}{2} – \sin^{-1}\left(\frac{1}{3}\right) \] \[ = \cos^{-1}\left(\frac{1}{3}\right) \] Thus, \[ x = \frac{1}{3} \] Final Answer: \[ \boxed{\frac{1}{3}} \]

If tan⁻¹(√3) + cot⁻¹x = π/2, find x Question If \[ \tan^{-1}(\sqrt{3}) + \cot^{-1}x = \frac{\pi}{2} \] Find \( x \). Solution We know: \[ \tan^{-1}(\sqrt{3}) = \frac{\pi}{3} \] So, \[ \frac{\pi}{3} + \cot^{-1}x = \frac{\pi}{2} \] \[ \cot^{-1}x = \frac{\pi}{2} – \frac{\pi}{3} = \frac{\pi}{6} \] Now, \[ \cot^{-1}x = \frac{\pi}{6} \Rightarrow x = \cot\left(\frac{\pi}{6}\right)

Value of sin⁻¹(sin 3π/5) Question Evaluate: \[ \sin^{-1}(\sin \tfrac{3\pi}{5}) \] Solution The principal value range of \( \sin^{-1}x \) is: \[ \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \] Now, \[ \frac{3\pi}{5} = 108^\circ \] Since \( \frac{3\pi}{5} \in \left(\frac{\pi}{2}, \pi\right) \), we use identity: \[ \sin^{-1}(\sin x) = \pi – x \quad \text{for } \frac{\pi}{2} < x < \pi

Prove sin⁻¹(2x√(1−x²)) = 2sin⁻¹x Question Show that: \[ \sin^{-1}\left(2x\sqrt{1-x^2}\right) = 2\sin^{-1}x \] Solution Let \[ \sin^{-1}x = \theta \] Then, \[ x = \sin \theta \] So, \[ \sqrt{1 – x^2} = \sqrt{1 – \sin^2\theta} = \cos \theta \] Now substitute in LHS: \[ \sin^{-1}\left(2x\sqrt{1-x^2}\right) = \sin^{-1}(2\sin\theta \cos\theta) \] Using identity: \[ 2\sin\theta \cos\theta =

Value of cos⁻¹(cos 5π/4) Question Find the value of: \[ \cos^{-1}(\cos \tfrac{5\pi}{4}) \] Solution The principal value range of \( \cos^{-1}x \) is: \[ [0, \pi] \] Now, \[ \frac{5\pi}{4} = \pi + \frac{\pi}{4} \] We use identity: \[ \cos(\pi + \theta) = -\cos \theta \] So, \[ \cos \tfrac{5\pi}{4} = -\frac{1}{\sqrt{2}} \] Now evaluate:

Value of tan⁻¹(a/b) − tan⁻¹((a−b)/(a+b)) Question Evaluate: \[ \tan^{-1}\left(\frac{a}{b}\right) – \tan^{-1}\left(\frac{a-b}{a+b}\right) \] Solution Use identity: \[ \tan^{-1}x – \tan^{-1}y = \tan^{-1}\left(\frac{x – y}{1 + xy}\right) \] Let \[ x = \frac{a}{b}, \quad y = \frac{a-b}{a+b} \] Then, \[ \frac{x – y}{1 + xy} = \frac{\frac{a}{b} – \frac{a-b}{a+b}}{1 + \frac{a}{b}\cdot \frac{a-b}{a+b}} \] Simplify numerator: \[ \frac{a(a+b)

Value of 2sin⁻¹(1/2) + cos⁻¹(−1/2) Question Find the value of: \[ 2\sin^{-1}\left(\frac{1}{2}\right) + \cos^{-1}\left(-\frac{1}{2}\right) \] Solution Using standard values: \[ \sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6} \] So, \[ 2\sin^{-1}\left(\frac{1}{2}\right) = 2 \cdot \frac{\pi}{6} = \frac{\pi}{3} \] Also, \[ \cos^{-1}\left(-\frac{1}{2}\right) = \frac{2\pi}{3} \] (since principal range of \( \cos^{-1}x \) is \( [0, \pi] \)) Therefore, \[ \frac{\pi}{3}

Value of tan⁻¹(tan 15π/4) Question Find the value of: \[ \tan^{-1}\left(\tan \frac{15\pi}{4}\right) \] Solution First, reduce the angle using periodicity of tangent: \[ \tan(\theta + \pi) = \tan \theta \] \[ \frac{15\pi}{4} = 3\pi + \frac{3\pi}{4} \] \[ \tan \frac{15\pi}{4} = \tan \frac{3\pi}{4} \] \[ \tan \frac{3\pi}{4} = -1 \] Now evaluate: \[ \tan^{-1}(-1) \]

Value of sin(π/3 − sin⁻¹(−1/2)) Question Find the value of: \[ \sin\left(\frac{\pi}{3} – \sin^{-1}\left(-\frac{1}{2}\right)\right) \] Solution First, evaluate: \[ \sin^{-1}\left(-\frac{1}{2}\right) = -\frac{\pi}{6} \] (since principal range of \( \sin^{-1}x \) is \( [-\frac{\pi}{2}, \frac{\pi}{2}] \)) Now substitute: \[ \sin\left(\frac{\pi}{3} – (-\frac{\pi}{6})\right) = \sin\left(\frac{\pi}{3} + \frac{\pi}{6}\right) \] \[ = \sin\left(\frac{\pi}{2}\right) \] \[ = 1 \] Final