Arts (English Medium) कक्षा १२ - CBSE Question Bank Solutions

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

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

If tan⁻¹ (cot θ) = 2 θ, then θ =

If \[\sin^{- 1} \left( \frac{2a}{1 - a^2} \right) + \cos^{- 1} \left( \frac{1 - a^2}{1 + a^2} \right) = \tan^{- 1} \left( \frac{2x}{1 - x^2} \right),\text{ where }a, x \in \left( 0, 1 \right)\] , then, the value of x is

The value of  \[\sin\left( 2\left( \tan^{- 1} 0 . 75 \right) \right)\] is equal to

If x > 1, then \[2 \tan^{- 1} x + \sin^{- 1} \left( \frac{2x}{1 + x^2} \right)\] is equal to

The domain of  \[\cos^{- 1} \left( x^2 - 4 \right)\] is

The value of \[\tan\left( \cos^{- 1} \frac{3}{5} + \tan^{- 1} \frac{1}{4} \right)\]

On expanding by first row, the value of the determinant of 3 × 3 square matrix
  \[A = \left[ a_{ij} \right]\text{ is }a_{11} C_{11} + a_{12} C_{12} + a_{13} C_{13}\] , where [C_ij] is the cofactor of a_ij in A. Write the expression for its value on expanding by second column.

Examine the continuity of the function  

\[f\left( x \right) = \left\{ \begin{array}{l}3x - 2, & x \leq 0 \\ x + 1 , & x > 0\end{array}at x = 0 \right.\]