April 2026

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

Evaluate cot(sin⁻¹(3/4) + sec⁻¹(4/3)) Problem Evaluate: \( \cot\left(\sin^{-1}\left(\frac{3}{4}\right) + \sec^{-1}\left(\frac{4}{3}\right)\right) \) Solution Let: \[ A = \sin^{-1}\left(\frac{3}{4}\right), \quad B = \sec^{-1}\left(\frac{4}{3}\right) \] Step 1: For A \[ \sin A = \frac{3}{4} = \frac{\text{Perpendicular}}{\text{Hypotenuse}} \] Perpendicular = 3 Hypotenuse = 4 Base: \[ \sqrt{4^2 – 3^2} = \sqrt{7} \] \[ \tan A = \frac{3}{\sqrt{7}} \] Step

Evaluate sin(cos⁻¹(−3/5) + cot⁻¹(−5/12)) Problem Evaluate: \( \sin\left(\cos^{-1}\left(\frac{-3}{5}\right) + \cot^{-1}\left(\frac{-5}{12}\right)\right) \) Solution Let: \[ A = \cos^{-1}\left(\frac{-3}{5}\right), \quad B = \cot^{-1}\left(\frac{-5}{12}\right) \] Step 1: Find sin A and cos A \[ \cos A = \frac{-3}{5} = \frac{\text{Base}}{\text{Hypotenuse}} \] Base = -3 Hypotenuse = 5 Perpendicular: \[ \sqrt{5^2 – 3^2} = \sqrt{25 – 9} = 4

Evaluate cos(tan⁻¹(−3/4)) Problem Evaluate: \( \cos\left(\tan^{-1}\left(\frac{-3}{4}\right)\right) \) Solution Let \( \theta = \tan^{-1}\left(\frac{-3}{4}\right) \) Then: \[ \tan \theta = \frac{-3}{4} = \frac{\text{Perpendicular}}{\text{Base}} \] Perpendicular = -3 Base = 4 Hypotenuse: \[ \sqrt{(-3)^2 + 4^2} = \sqrt{9 + 16} = 5 \] Now, \[ \cos \theta = \frac{\text{Base}}{\text{Hypotenuse}} = \frac{4}{5} \] Therefore: \[ \cos\left(\tan^{-1}\left(\frac{-3}{4}\right)\right) = \frac{4}{5}

Evaluate cosec(cot⁻¹(−12/5)) Problem Evaluate: \( \csc\left(\cot^{-1}\left(\frac{-12}{5}\right)\right) \) Solution Let \( \theta = \cot^{-1}\left(\frac{-12}{5}\right) \) Then: \[ \cot \theta = \frac{-12}{5} = \frac{\text{Base}}{\text{Perpendicular}} \] Base = -12 Perpendicular = 5 Hypotenuse: \[ \sqrt{(-12)^2 + 5^2} = \sqrt{144 + 25} = 13 \] Now, \[ \sin \theta = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{5}{13} \] \[ \csc \theta = \frac{1}{\sin

Evaluate tan(cos⁻¹(−7/25)) Problem Evaluate: \( \tan\left(\cos^{-1}\left(\frac{-7}{25}\right)\right) \) Solution Let \( \theta = \cos^{-1}\left(\frac{-7}{25}\right) \) Then: \[ \cos \theta = \frac{-7}{25} = \frac{\text{Base}}{\text{Hypotenuse}} \] Base = -7 Hypotenuse = 25 Perpendicular: \[ \sqrt{25^2 – 7^2} = \sqrt{625 – 49} = \sqrt{576} = 24 \] Now, \[ \tan \theta = \frac{\text{Perpendicular}}{\text{Base}} = \frac{24}{-7} \] Therefore: \[ \tan\left(\cos^{-1}\left(\frac{-7}{25}\right)\right)

Evaluate cot(sec⁻¹(−13/5)) Problem Evaluate: \( \cot\left(\sec^{-1}\left(\frac{-13}{5}\right)\right) \) Solution Let \( \theta = \sec^{-1}\left(\frac{-13}{5}\right) \) Then: \[ \sec \theta = \frac{-13}{5} \] \[ \cos \theta = \frac{1}{\sec \theta} = -\frac{5}{13} \] Using triangle: Base = -5 Hypotenuse = 13 Perpendicular: \[ \sqrt{13^2 – 5^2} = \sqrt{169 – 25} = 12 \] Now, \[ \cot \theta =

Evaluate sec(cot⁻¹(−5/12)) Problem Evaluate: \( \sec\left(\cot^{-1}\left(\frac{-5}{12}\right)\right) \) Solution Let \( \theta = \cot^{-1}\left(\frac{-5}{12}\right) \) Then: \[ \cot \theta = \frac{-5}{12} = \frac{\text{Base}}{\text{Perpendicular}} \] Base = -5 Perpendicular = 12 Hypotenuse: \[ \sqrt{(-5)^2 + 12^2} = \sqrt{25 + 144} = 13 \] Now, \[ \cos \theta = \frac{\text{Base}}{\text{Hypotenuse}} = \frac{-5}{13} \] \[ \sec \theta = \frac{1}{\cos