Ravi Kant Kumar

If x < 0, y < 0 and xy = 1, find tan⁻¹x + tan⁻¹y Question If \( x < 0 \), \( y < 0 \) and \( xy = 1 \), find: \[ \tan^{-1}x + \tan^{-1}y \] Solution We use identity: \[ \tan^{-1}x + \tan^{-1}y = \tan^{-1}\left(\frac{x+y}{1-xy}\right) \] Given: \[ xy = 1 […]

Number of solutions of √(1+cos2x) = √2 sin⁻¹(sin x) Question Find the number of real solutions of: \[ \sqrt{1+\cos 2x} = \sqrt{2}\,\sin^{-1}(\sin x), \quad -\pi \le x \le \pi \] Solution Simplify LHS: \[ 1 + \cos 2x = 2\cos^2 x \] \[ \sqrt{1+\cos 2x} = \sqrt{2\cos^2 x} = \sqrt{2}\,|\cos x| \] So equation becomes:

Relation between α and β Question If \[ \alpha = \tan^{-1}(\tan \tfrac{5\pi}{4}), \quad \beta = \tan^{-1}(-\tan \tfrac{2\pi}{3}) \] Find the relation between \( \alpha \) and \( \beta \). Solution Principal value range of \( \tan^{-1}x \) is: \[ \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \] Find α: \[ \tan \tfrac{5\pi}{4} = 1 \Rightarrow \alpha = \tan^{-1}(1) = \frac{\pi}{4}

Number of solutions of tan⁻¹(2x) + tan⁻¹(3x) = π/4 Question Find the number of solutions of: \[ \tan^{-1}(2x) + \tan^{-1}(3x) = \frac{\pi}{4} \] Solution Use identity: \[ \tan^{-1}a + \tan^{-1}b = \tan^{-1}\left(\frac{a+b}{1-ab}\right) \] So, \[ \tan^{-1}\left(\frac{2x + 3x}{1 – 6x^2}\right) = \frac{\pi}{4} \] \[ \tan^{-1}\left(\frac{5x}{1 – 6x^2}\right) = \frac{\pi}{4} \] Thus, \[ \frac{5x}{1 – 6x^2}

Value of sin[cot⁻¹{tan(cos⁻¹x)}] Question Simplify: \[ \sin\left[\cot^{-1}\{\tan(\cos^{-1}x)\}\right] \] Solution Let \[ \cos^{-1}x = \theta \Rightarrow \cos\theta = x \] Then, \[ \tan(\cos^{-1}x) = \tan\theta \] Now, \[ \sin\left(\cot^{-1}(\tan\theta)\right) \] Let \[ \cot^{-1}(\tan\theta) = \phi \Rightarrow \cot\phi = \tan\theta \Rightarrow \tan\phi = \cot\theta \] So, \[ \phi = \frac{\pi}{2} – \theta \] Thus, \[ \sin\phi =

If sin⁻¹x − cos⁻¹x = π/6, find x Question If \[ \sin^{-1}x – \cos^{-1}x = \frac{\pi}{6} \] Find \( x \). Solution We use identity: \[ \sin^{-1}x + \cos^{-1}x = \frac{\pi}{2} \] Let \( \sin^{-1}x = \theta \) Then: \[ \theta – \left(\frac{\pi}{2} – \theta\right) = \frac{\pi}{6} \] \[ 2\theta – \frac{\pi}{2} = \frac{\pi}{6} \]

Positive integral solution of given inverse trig equation Question Find the positive integral solution of: \[ \tan^{-1}x + \cos^{-1}\left(\frac{y}{\sqrt{1+y^2}}\right) = \sin^{-1}\left(\frac{3}{\sqrt{10}}\right) \] Solution We know identity: \[ \cos^{-1}\left(\frac{y}{\sqrt{1+y^2}}\right) = \tan^{-1}\left(\frac{1}{y}\right) \] So equation becomes: \[ \tan^{-1}x + \tan^{-1}\left(\frac{1}{y}\right) = \sin^{-1}\left(\frac{3}{\sqrt{10}}\right) \] Now, \[ \sin^{-1}\left(\frac{3}{\sqrt{10}}\right) = \tan^{-1}(3) \] (since \( \sin\theta = 3/\sqrt{10} \Rightarrow \tan\theta =

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 –