Commerce (English Medium) कक्षा ११ - CBSE Question Bank Solutions for Mathematics

The value of \[\cos \frac{\pi}{65} \cos \frac{2\pi}{65} \cos \frac{4\pi}{65} \cos \frac{8\pi}{65} \cos \frac{16\pi}{65} \cos \frac{32\pi}{65}\]  is 

If \[\cos 2x + 2 \cos x = 1\]  then, \[\left( 2 - \cos^2 x \right) \sin^2 x\]  is equal to 

For all real values of x, \[\cot x - 2 \cot 2x\] is equal to 

The value of  \[2 \tan \frac{\pi}{10} + 3 \sec \frac{\pi}{10} - 4 \cos \frac{\pi}{10}\] is 

If in a  \[∆ ABC, \tan A + \tan B + \tan C = 0\], then

If \[\cos x = \frac{1}{2} \left( a + \frac{1}{a} \right),\]  and \[\cos 3 x = \lambda \left( a^3 + \frac{1}{a^3} \right)\] then \[\lambda =\]

If  \[2 \tan \alpha = 3 \tan \beta, \text{ then }  \tan \left( \alpha - \beta \right) =\]

If \[\tan \alpha = \frac{1 - \cos \beta}{\sin \beta}\] , then

If \[\sin \alpha + \sin \beta = a \text{ and }  \cos \alpha - \cos \beta = b \text{ then }  \tan \frac{\alpha - \beta}{2} =\]

The value of \[\left( \cot \frac{x}{2} - \tan \frac{x}{2} \right)^2 \left( 1 - 2 \tan x \cot 2 x \right)\] is 

The value of \[\tan x \sin \left( \frac{\pi}{2} + x \right) \cos \left( \frac{\pi}{2} - x \right)\]

\[\sin^2 \left( \frac{\pi}{18} \right) + \sin^2 \left( \frac{\pi}{9} \right) + \sin^2 \left( \frac{7\pi}{18} \right) + \sin^2 \left( \frac{4\pi}{9} \right) =\]

If  \[5 \sin \alpha = 3 \sin \left( \alpha + 2 \beta \right) \neq 0\] , then \[\tan \left( \alpha + \beta \right)\]  is equal to

\[2 \text{ cos } x - \ cos  3x - \cos 5x - 16 \cos^3 x \sin^2 x\]

If \[A = 2 \sin^2 x - \cos 2x\] , then A lies in the interval

The value of \[\frac{\cos 3x}{2 \cos 2x - 1}\]  is equal to

If \[\tan \left( \pi/4 + x \right) + \tan \left( \pi/4 - x \right) = \lambda \sec 2x, \text{ then } \]

The value of  \[\cos^2 \left( \frac{\pi}{6} + x \right) - \sin^2 \left( \frac{\pi}{6} - x \right)\] is 

\[\frac{\sin 3x}{1 + 2 \cos 2x}\]   is equal to