Ravi Kant Kumar

Value of sin⁻¹(cos π/9) Question Find the value of: \[ \sin^{-1}(\cos \tfrac{\pi}{9}) \] Solution Use identity: \[ \cos \theta = \sin\left(\frac{\pi}{2} – \theta\right) \] So, \[ \sin^{-1}(\cos \tfrac{\pi}{9}) = \sin^{-1}\left(\sin\left(\frac{\pi}{2} – \frac{\pi}{9}\right)\right) \] \[ = \sin^{-1}\left(\sin \tfrac{7\pi}{18}\right) \] Now check principal range of \( \sin^{-1}x \): \[ \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \] Since \( \tfrac{7\pi}{18} \in \left(0, […]

Value of cos⁻¹(cos 6) Question Find the value of: \[ \cos^{-1}(\cos 6) \] Solution We know the principal value range of \( \cos^{-1}x \) is: \[ [0, \pi] \] Now, note that: \[ 2\pi \approx 6.283 \Rightarrow 6 = 2\pi – 0.283 \] So, \[ \cos 6 = \cos(2\pi – 0.283) = \cos(0.283) \] Thus,

If tan⁻¹x + tan⁻¹y = π/4, find relation between x and y Question If \[ \tan^{-1}x + \tan^{-1}y = \frac{\pi}{4} \] Find the relation between \( x \) and \( y \). Solution Take tangent on both sides: \[ \tan\left(\tan^{-1}x + \tan^{-1}y\right) = \tan\frac{\pi}{4} \] Using identity: \[ \tan(A+B) = \frac{\tan A + \tan B}{1

Value of cos²(½ cos⁻¹(3/5)) Question Find the value of: \[ \cos^2\left(\frac{1}{2}\cos^{-1}\left(\frac{3}{5}\right)\right) \] Solution Let \[ \theta = \cos^{-1}\left(\frac{3}{5}\right) \] Then, \[ \cos \theta = \frac{3}{5} \] Using identity: \[ \cos^2\left(\frac{\theta}{2}\right) = \frac{1 + \cos \theta}{2} \] Substitute: \[ \cos^2\left(\frac{\theta}{2}\right) = \frac{1 + \frac{3}{5}}{2} \] \[ = \frac{\frac{8}{5}}{2} = \frac{8}{10} = \frac{4}{5} \] Final Answer: \[

Value of cos⁻¹(cos 350°) − sin⁻¹(sin 350°) Question Find the value of: \[ \cos^{-1}(\cos 350^\circ) – \sin^{-1}(\sin 350^\circ) \] Solution We use principal value ranges: \( \cos^{-1}x \in [0^\circ, 180^\circ] \) \( \sin^{-1}x \in [-90^\circ, 90^\circ] \) First, \[ \cos^{-1}(\cos 350^\circ) \] Since \( 350^\circ = 360^\circ – 10^\circ \), \[ \cos 350^\circ = \cos

Value of cos(2sin⁻¹(1/2)) Question Find the value of: \[ \cos\left(2\sin^{-1}\left(\frac{1}{2}\right)\right) \] Solution Let \[ \theta = \sin^{-1}\left(\frac{1}{2}\right) \] Then, \[ \sin \theta = \frac{1}{2} \Rightarrow \theta = \frac{\pi}{6} \] Now, \[ \cos(2\theta) = \cos\left(2 \cdot \frac{\pi}{6}\right) = \cos\left(\frac{\pi}{3}\right) \] \[ = \frac{1}{2} \] Final Answer: \[ \boxed{\frac{1}{2}} \] Key Concept Use known standard values to

Value of cos⁻¹(tan 3π/4) Question Find the value of: \[ \cos^{-1}(\tan \tfrac{3\pi}{4}) \] Solution First, evaluate: \[ \tan \tfrac{3\pi}{4} = -1 \] So the expression becomes: \[ \cos^{-1}(-1) \] We know that: \[ \cos \pi = -1 \] And the principal value range of \( \cos^{-1}x \) is: \[ [0, \pi] \] Thus, \[ \cos^{-1}(-1)

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 \(