Ravi Kant Kumar

Solve sin⁻¹x = π/6 + cos⁻¹x Problem Solve: \( \sin^{-1}x = \frac{\pi}{6} + \cos^{-1}x \) Solution Step 1: Use identity \[ \cos^{-1}x = \frac{\pi}{2} – \sin^{-1}x \] Step 2: Substitute \[ \sin^{-1}x = \frac{\pi}{6} + \left(\frac{\pi}{2} – \sin^{-1}x\right) \] \[ \sin^{-1}x = \frac{2\pi}{3} – \sin^{-1}x \] Step 3: Solve \[ 2\sin^{-1}x = \frac{2\pi}{3} \] \[ […]

Solve sin(sin⁻¹(1/5) + cos⁻¹x) = 1 Problem Solve: \( \sin\left(\sin^{-1}\left(\frac{1}{5}\right) + \cos^{-1}(x)\right) = 1 \) Solution Step 1: Use property of sine \[ \sin \theta = 1 \Rightarrow \theta = \frac{\pi}{2} \] So, \[ \sin^{-1}\left(\frac{1}{5}\right) + \cos^{-1}(x) = \frac{\pi}{2} \] Step 2: Convert using identity \[ \cos^{-1}x = \frac{\pi}{2} – \sin^{-1}x \] Step 3: Substitute

Find x in (sin⁻¹x)² + (cos⁻¹x)² = 17π²/36 Problem Solve: \( (\sin^{-1}x)^2 + (\cos^{-1}x)^2 = \frac{17\pi^2}{36} \) Solution Step 1: Use identity \[ \sin^{-1}x + \cos^{-1}x = \frac{\pi}{2} \] Let \( a = \sin^{-1}x \), then: \[ \cos^{-1}x = \frac{\pi}{2} – a \] Step 2: Substitute \[ a^2 + \left(\frac{\pi}{2} – a\right)^2 = \frac{17\pi^2}{36} \]

Find x in cot(cos⁻¹(3/5) + sin⁻¹(x)) = 0 Problem Find x if \( \cot\left(\cos^{-1}\left(\frac{3}{5}\right) + \sin^{-1}(x)\right) = 0 \) Solution Step 1: Use property of cot \[ \cot \theta = 0 \Rightarrow \theta = \frac{\pi}{2} \] So, \[ \cos^{-1}\left(\frac{3}{5}\right) + \sin^{-1}(x) = \frac{\pi}{2} \] Step 2: Convert using identity \[ \cos^{-1}t = \frac{\pi}{2} – \sin^{-1}t

Find x and y Problem If \( \sin^{-1}(x) + \sin^{-1}(y) = \frac{\pi}{3} \) and \( \cos^{-1}(x) – \cos^{-1}(y) = \frac{\pi}{6} \), find x and y. Solution Step 1: Convert cos⁻¹ into sin⁻¹ \[ \cos^{-1}t = \frac{\pi}{2} – \sin^{-1}t \] So, \[ \cos^{-1}x – \cos^{-1}y = \left(\frac{\pi}{2} – \sin^{-1}x\right) – \left(\frac{\pi}{2} – \sin^{-1}y\right) \] \[ =

Find sin⁻¹x + sin⁻¹y Problem If \( \cos^{-1}x + \cos^{-1}y = \frac{\pi}{4} \), find \( \sin^{-1}x + \sin^{-1}y \). Solution Step 1: Use identity \[ \sin^{-1}t + \cos^{-1}t = \frac{\pi}{2} \] So, \[ \sin^{-1}x = \frac{\pi}{2} – \cos^{-1}x \] \[ \sin^{-1}y = \frac{\pi}{2} – \cos^{-1}y \] Step 2: Add both \[ \sin^{-1}x + \sin^{-1}y =

Evaluate cos(sec⁻¹x + cosec⁻¹x) Problem Evaluate: \( \cos\left(\sec^{-1}x + \csc^{-1}x\right), \quad |x| \ge 1 \) Solution Let: \[ A = \sec^{-1}x,\quad B = \csc^{-1}x \] Step 1: Convert to sine and cosine \[ \sec A = x \Rightarrow \cos A = \frac{1}{x} \] \[ \csc B = x \Rightarrow \sin B = \frac{1}{x} \] Step

Evaluate cot(tan⁻¹(a) + cot⁻¹(a)) Problem Evaluate: \( \cot\left(\tan^{-1}(a) + \cot^{-1}(a)\right) \) Solution Use identity: \[ \tan^{-1}(a) + \cot^{-1}(a) = \frac{\pi}{2} \] Therefore: \[ \cot\left(\tan^{-1}(a) + \cot^{-1}(a)\right) = \cot\left(\frac{\pi}{2}\right) \] \[ = 0 \] Final Answer \[ \boxed{0} \] Explanation The sum of tan⁻¹(a) and cot⁻¹(a) is always π/2, so cot(π/2) = 0. Next Question /

Evaluate sin(tan⁻¹x + tan⁻¹(1/x)) for x > 0

Evaluate sin(tan⁻¹x + tan⁻¹(1/x)) for x < 0 Problem Evaluate: \( \sin\left(\tan^{-1}x + \tan^{-1}\frac{1}{x}\right), \quad x < 0 \) Solution Let: \[ A = \tan^{-1}x,\quad B = \tan^{-1}\frac{1}{x} \] Step 1: Use identity \[ \tan^{-1}x + \tan^{-1}\frac{1}{x} = \begin{cases} \frac{\pi}{2}, & x > 0 \\ -\frac{\pi}{2}, & x < 0 \end{cases} \] Since \( x