Ravi Kant Kumar

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

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

Solve cos(2sin⁻¹(−x)) = 0 Problem Solve: \( \cos\left(2\sin^{-1}(-x)\right) = 0 \) Solution Use identity: \[ \cos(2\theta) = 1 – 2\sin^2\theta \] Let \( \theta = \sin^{-1}(-x) \) Then: \[ \sin \theta = -x \] Now, \[ \cos\left(2\sin^{-1}(-x)\right) = 1 – 2(-x)^2 \] \[ = 1 – 2x^2 \] Given: \[ 1 – 2x^2 = 0

Solve cos(sin⁻¹x) = 1/6 Problem Solve: \( \cos(\sin^{-1}x) = \frac{1}{6} \) Solution Let \( \theta = \sin^{-1}x \) Then: \[ \sin \theta = x = \frac{\text{Perpendicular}}{\text{Hypotenuse}} \] So take: Perpendicular = x Hypotenuse = 1 Base: \[ \sqrt{1 – x^2} \] Now, \[ \cos \theta = \frac{\text{Base}}{\text{Hypotenuse}} = \sqrt{1 – x^2} \] Given: \[ \sqrt{1