April 2026

Find expression from cos⁻¹(x/a) + cos⁻¹(y/b) = α Question If \[ \cos^{-1}\left(\frac{x}{a}\right) + \cos^{-1}\left(\frac{y}{b}\right) = \alpha \] Find: \[ \frac{x^2}{a^2} – \frac{2xy}{ab}\cos\alpha + \frac{y^2}{b^2} \] Solution Let \[ \cos^{-1}\left(\frac{x}{a}\right) = A,\quad \cos^{-1}\left(\frac{y}{b}\right) = B \] Then, \[ A + B = \alpha \] So, \[ \cos A = \frac{x}{a}, \quad \cos B = \frac{y}{b} \] […]

Value of 2tan⁻¹{cosec(tan⁻¹x) − tan(cot⁻¹x)} Question Evaluate: \[ 2\tan^{-1}\left\{\csc(\tan^{-1}x) – \tan(\cot^{-1}x)\right\} \] Solution Let \[ \tan^{-1}x = \theta \Rightarrow \tan\theta = x \] Then, \[ \sin\theta = \frac{x}{\sqrt{1+x^2}}, \quad \csc\theta = \frac{\sqrt{1+x^2}}{x} \] So, \[ \csc(\tan^{-1}x) = \frac{\sqrt{1+x^2}}{x} \] Now let \[ \cot^{-1}x = \phi \Rightarrow \cot\phi = x \Rightarrow \tan\phi = \frac{1}{x} \] So,

Value of tan{cos⁻¹(1/(5√2)) − sin⁻¹(4/√17)} Question Evaluate: \[ \tan\left(\cos^{-1}\left(\frac{1}{5\sqrt{2}}\right) – \sin^{-1}\left(\frac{4}{\sqrt{17}}\right)\right) \] Solution Let \[ A = \cos^{-1}\left(\frac{1}{5\sqrt{2}}\right), \quad B = \sin^{-1}\left(\frac{4}{\sqrt{17}}\right) \] Use identity: \[ \tan(A – B) = \frac{\tan A – \tan B}{1 + \tan A \tan B} \] Find tan A: \[ \cos A = \frac{1}{5\sqrt{2}} \Rightarrow \sin A = \sqrt{1 –

Find x² from tan⁻¹ expression Question If \[ \tan^{-1}\left(\frac{\sqrt{1+x^2} – \sqrt{1-x^2}}{\sqrt{1+x^2} + \sqrt{1-x^2}}\right) = \alpha \] Find \( x^2 \). Solution Let \[ \tan^{-1}(A) = \alpha \Rightarrow A = \tan \alpha \] So, \[ \tan \alpha = \frac{\sqrt{1+x^2} – \sqrt{1-x^2}}{\sqrt{1+x^2} + \sqrt{1-x^2}} \] This is a standard identity: \[ \tan\left(\frac{\theta}{2}\right) = \frac{1 – \cos \theta}{\sin

Value of tan⁻¹(tan 9π/8) Question Find the value of: \[ \tan^{-1}(\tan \tfrac{9\pi}{8}) \] Solution The principal value range of \( \tan^{-1}x \) is: \[ \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \] Now reduce the angle: \[ \frac{9\pi}{8} = \pi + \frac{\pi}{8} \Rightarrow \tan \tfrac{9\pi}{8} = \tan \tfrac{\pi}{8} \] But \( \tfrac{\pi}{8} \in \left(0, \frac{\pi}{2}\right) \), so \[ \tan^{-1}(\tan \tfrac{\pi}{8})

Value of cos⁻¹(cos 13π/6) Question Find the value of: \[ \cos^{-1}(\cos \tfrac{13\pi}{6}) \] Solution First, reduce the angle: \[ \frac{13\pi}{6} = 2\pi + \frac{\pi}{6} \Rightarrow \cos \tfrac{13\pi}{6} = \cos \tfrac{\pi}{6} \] Now evaluate: \[ \cos^{-1}(\cos \tfrac{\pi}{6}) \] The principal value range of \( \cos^{-1}x \) is: \[ [0, \pi] \] Since \( \tfrac{\pi}{6} \in [0,

If cos(sin⁻¹(2/5) + cos⁻¹x) = 0, find x Question If \[ \cos\left(\sin^{-1}\left(\frac{2}{5}\right) + \cos^{-1}x\right) = 0 \] Find \( x \). Solution We know: \[ \cos \theta = 0 \Rightarrow \theta = \frac{\pi}{2} \quad \text{(principal value)} \] So, \[ \sin^{-1}\left(\frac{2}{5}\right) + \cos^{-1}x = \frac{\pi}{2} \] Using identity: \[ \sin^{-1}a + \cos^{-1}a = \frac{\pi}{2} \] Comparing,

Value of 2sec⁻¹(2) + sin⁻¹(1/2) Question Evaluate: \[ 2\sec^{-1}(2) + \sin^{-1}\left(\frac{1}{2}\right) \] Solution We know: \[ \sec^{-1}(2) = \cos^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3} \] So, \[ 2\sec^{-1}(2) = 2 \cdot \frac{\pi}{3} = \frac{2\pi}{3} \] Also, \[ \sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6} \] Therefore, \[ \frac{2\pi}{3} + \frac{\pi}{6} \] \[ = \frac{4\pi}{6} + \frac{\pi}{6} = \frac{5\pi}{6} \] Final Answer: \[

Value of cos((tan⁻¹x + cot⁻¹x)/3) when x = −1/√3 Question Evaluate: \[ \cos\left(\frac{\tan^{-1}x + \cot^{-1}x}{3}\right) \quad \text{when } x = -\frac{1}{\sqrt{3}} \] Solution We use identity: \[ \tan^{-1}x + \cot^{-1}x = \frac{\pi}{2} \quad \text{for all } x \in \mathbb{R} \] So, \[ \cos\left(\frac{\pi/2}{3}\right) = \cos\left(\frac{\pi}{6}\right) \] \[ = \frac{\sqrt{3}}{2} \] Final Answer: \[ \boxed{\frac{\sqrt{3}}{2}} \]

If cos(tan⁻¹x + cot⁻¹√3) = 0, find x Question If \[ \cos\left(\tan^{-1}x + \cot^{-1}\sqrt{3}\right) = 0 \] Find \( x \). Solution We know: \[ \cot^{-1}\sqrt{3} = \frac{\pi}{6} \] So the equation becomes: \[ \cos\left(\tan^{-1}x + \frac{\pi}{6}\right) = 0 \] Cosine is zero when: \[ \theta = \frac{\pi}{2} \quad \text{(within principal consideration)} \] Thus, \[