Ravi Kant Kumar

Evaluate sin(tan⁻¹(3/4)) Question Evaluate: \[ \sin(\tan^{-1}\left(\frac{3}{4}\right)) \] Solution Let \[ \theta = \tan^{-1}\left(\frac{3}{4}\right) \] Then, \[ \tan \theta = \frac{3}{4} = \frac{\text{Opposite}}{\text{Adjacent}} \] Take a right triangle: Opposite = 3 Adjacent = 4 Hypotenuse: \[ \sqrt{3^2 + 4^2} = 5 \] Now, \[ \sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{3}{5} \] Final Answer: \[ \boxed{\frac{3}{5}} \] […]

Evaluate sin(½ cos⁻¹(4/5)) Question Evaluate: \[ \sin\left(\frac{1}{2}\cos^{-1}\left(\frac{4}{5}\right)\right) \] Solution Let \[ \theta = \cos^{-1}\left(\frac{4}{5}\right) \] Then, \[ \cos \theta = \frac{4}{5} \] Construct a right triangle: Adjacent = 4 Hypotenuse = 5 Opposite = 3 So, \[ \sin \theta = \frac{3}{5} \] Now use half-angle identity: \[ \sin\frac{\theta}{2} = \sqrt{\frac{1 – \cos \theta}{2}} \] Substitute:

Value of sin⁻¹(sin 1550°) Question Find the value of: \[ \sin^{-1}(\sin 1550^\circ) \] Solution First, reduce the angle: \[ 1550^\circ = 4 \times 360^\circ + 110^\circ \] \[ \sin 1550^\circ = \sin 110^\circ \] Now evaluate: \[ \sin^{-1}(\sin 110^\circ) \] The principal value range of \( \sin^{-1}x \) is: \[ [-90^\circ, 90^\circ] \] Since \(

Value of cos(2sin⁻¹(1/3)) Question Find the value of: \[ \cos\left(2\sin^{-1}\left(\frac{1}{3}\right)\right) \] Solution Let \[ \theta = \sin^{-1}\left(\frac{1}{3}\right) \] Then, \[ \sin \theta = \frac{1}{3} \] Using identity: \[ \cos 2\theta = 1 – 2\sin^2\theta \] Substitute: \[ \cos 2\theta = 1 – 2\left(\frac{1}{3}\right)^2 \] \[ = 1 – 2 \cdot \frac{1}{9} = 1 – \frac{2}{9}

Value of sin⁻¹(sin −600°) Question Find the value of: \[ \sin^{-1}(\sin (-600^\circ)) \] Solution First, reduce the angle: \[ -600^\circ = -600^\circ + 720^\circ = 120^\circ \] \[ \sin(-600^\circ) = \sin(120^\circ) \] Now evaluate: \[ \sin^{-1}(\sin 120^\circ) \] The principal value range of \( \sin^{-1}x \) is: \[ [-90^\circ, 90^\circ] \] Since \( 120^\circ \)

Value of cos⁻¹(cos 1540°) Question Find the value of: \[ \cos^{-1}(\cos 1540^\circ) \] Solution First, reduce the angle using periodicity: \[ 1540^\circ = 4 \times 360^\circ + 100^\circ \] \[ \cos 1540^\circ = \cos 100^\circ \] Now, we evaluate: \[ \cos^{-1}(\cos 100^\circ) \] The principal value range of \( \cos^{-1}x \) is: \[ [0^\circ, 180^\circ]

Range of tan⁻¹x Question Write the range of: \[ \tan^{-1}x \] Solution The inverse tangent function \( \tan^{-1}x \) is defined for all real values of \( x \). Its principal value range is: \[ \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \] This means: \( \tan^{-1}x \) never equals \( \frac{\pi}{2} \) or \( -\frac{\pi}{2} \) But it approaches

Value of cos⁻¹(1/2) + 2sin⁻¹(1/2) Question Find the value of: \[ \cos^{-1}\left(\frac{1}{2}\right) + 2\sin^{-1}\left(\frac{1}{2}\right) \] Solution Using standard values: \[ \cos^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3} \] (since \( \cos \frac{\pi}{3} = \frac{1}{2} \) and lies in principal range) Also, \[ \sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6} \] (since \( \sin \frac{\pi}{6} = \frac{1}{2} \)) Therefore, \[ \cos^{-1}\left(\frac{1}{2}\right) + 2\sin^{-1}\left(\frac{1}{2}\right) =

Value of sin(cot⁻¹x) Question Find the value of: \[ \sin(\cot^{-1}x) \] Solution Let \[ \theta = \cot^{-1}x \] Then, \[ \cot \theta = x = \frac{\text{Adjacent}}{\text{Opposite}} \] Take a right triangle such that: Adjacent = \( x \) Opposite = \( 1 \) Then hypotenuse: \[ \sqrt{x^2 + 1} \] Now, \[ \sin \theta =

If −1 < x < 0, find sin⁻¹(2x/(1+x²)) + cos⁻¹((1−x²)/(1+x²)) Question If \( -1 < x < 0 \), find the value of: \[ \sin^{-1}\left(\frac{2x}{1+x^2}\right) + \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) \] Solution Let \[ x = \tan \theta \] Then, since \( -1 < x < 0 \Rightarrow \theta \in \left(-\frac{\pi}{4}, 0\right) \) Using identities: \[ \sin 2\theta