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

Find the value of \[2 \sec^{- 1} 2 + \sin^{- 1} \left( \frac{1}{2} \right)\]

If \[\cos\left( \sin^{- 1} \frac{2}{5} + \cos^{- 1} x \right) = 0\], find the value of x.

Find the value of \[\cos^{- 1} \left( \cos\frac{13\pi}{6} \right)\]

Find the value of \[\tan^{- 1} \left( \tan\frac{9\pi}{8} \right)\]

If \[\tan^{- 1} \left( \frac{\sqrt{1 + x^2} - \sqrt{1 - x^2}}{\sqrt{1 + x^2} + \sqrt{1 - x^2}} \right)\]  = α, then x² =

The value of tan \[\left\{ \cos^{- 1} \frac{1}{5\sqrt{2}} - \sin^{- 1} \frac{4}{\sqrt{17}} \right\}\] is

2 tan⁻¹ {cosec (tan⁻¹ x) − tan (cot⁻¹ x)} is equal to

If  \[\cos^{- 1} \frac{x}{a} + \cos^{- 1} \frac{y}{b} = \alpha, then\frac{x^2}{a^2} - \frac{2xy}{ab}\cos \alpha + \frac{y^2}{b^2} = \]

The positive integral solution of the equation
\[\tan^{- 1} x + \cos^{- 1} \frac{y}{\sqrt{1 + y^2}} = \sin^{- 1} \frac{3}{\sqrt{10}}\text{ is }\]

If sin⁻¹ x − cos⁻¹ x = `pi/6` , then x = 

sin\[\left[ \cot^{- 1} \left\{ \tan\left( \cos^{- 1} x \right) \right\} \right]\]  is equal to

The number of solutions of the equation \[\tan^{- 1} 2x + \tan^{- 1} 3x = \frac{\pi}{4}\] is

If α = \[\tan^{- 1} \left( \tan\frac{5\pi}{4} \right) \text{ and }\beta = \tan^{- 1} \left( - \tan\frac{2\pi}{3} \right)\] , then

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