Science (English Medium) कक्षा १२ - CBSE Question Bank Solutions for Mathematics

The number of real solutions of the equation \[\sqrt{1 + \cos 2x} = \sqrt{2} \sin^{- 1} (\sin x), - \pi \leq x \leq \pi\]

If x < 0, y < 0 such that xy = 1, then tan⁻¹ x + tan⁻¹ y equals

\[\text{ If } u = \cot^{- 1} \sqrt{\tan \theta} - \tan^{- 1} \sqrt{\tan \theta}\text{ then }, \tan\left( \frac{\pi}{4} - \frac{u}{2} \right) =\]

\[\text{ If }\cos^{- 1} \frac{x}{3} + \cos^{- 1} \frac{y}{2} = \frac{\theta}{2}, \text{ then }4 x^2 - 12xy \cos\frac{\theta}{2} + 9 y^2 =\]

If α = \[\tan^{- 1} \left( \frac{\sqrt{3}x}{2y - x} \right), \beta = \tan^{- 1} \left( \frac{2x - y}{\sqrt{3}y} \right),\] 
 then α − β =

Let f (x) = `e^(cos^-1){sin(x+pi/3}.`
Then, f (8π/9) = 

\[\tan^{- 1} \frac{1}{11} + \tan^{- 1} \frac{2}{11}\]  is equal to

If \[\cos^{- 1} \frac{x}{2} + \cos^{- 1} \frac{y}{3} = \theta,\]  then 9x² − 12xy cos θ + 4y² is equal to

If tan⁻¹ 3 + tan⁻¹ x = tan⁻¹ 8, then x =

The value of \[\sin^{- 1} \left( \cos\frac{33\pi}{5} \right)\] is 

The value of \[\cos^{- 1} \left( \cos\frac{5\pi}{3} \right) + \sin^{- 1} \left( \sin\frac{5\pi}{3} \right)\] is

sin \[\left\{ 2 \cos^{- 1} \left( \frac{- 3}{5} \right) \right\}\]  is equal to

If θ = sin⁻¹ {sin (−600°)}, then one of the possible values of θ is

If \[3\sin^{- 1} \left( \frac{2x}{1 + x^2} \right) - 4 \cos^{- 1} \left( \frac{1 - x^2}{1 + x^2} \right) + 2 \tan^{- 1} \left( \frac{2x}{1 - x^2} \right) = \frac{\pi}{3}\] is equal to

If 4 cos⁻¹ x + sin⁻¹ x = π, then the value of x is

It \[\tan^{- 1} \frac{x + 1}{x - 1} + \tan^{- 1} \frac{x - 1}{x} = \tan^{- 1}\]   (−7), then the value of x is

If \[\cos^{- 1} x > \sin^{- 1} x\], then

In a ∆ ABC, if C is a right angle, then
\[\tan^{- 1} \left( \frac{a}{b + c} \right) + \tan^{- 1} \left( \frac{b}{c + a} \right) =\]

The value of sin \[\left( \frac{1}{4} \sin^{- 1} \frac{\sqrt{63}}{8} \right)\] is

\[\cot\left( \frac{\pi}{4} - 2 \cot^{- 1} 3 \right) =\]