If $\sin A + \cos A = \sqrt 2 ,$ then ${\cos ^2}A = $
If $\sin A + \cos A = \sqrt 2 ,$ then ${\cos ^2}A = $
- A
$\frac{1}{4}$
- B
$\frac{1}{2}$
- C
$\frac{1}{{\sqrt 2 }}$
- D
$\frac{3}{2}$
Similar Questions
If $\tan x = \frac{{2b}}{{a - c}}(a \ne c),$
$y = a\,{\cos ^2}x + 2b\,\sin x\cos x + c\,{\sin ^2}x$
and $z = a{\sin ^2}x - 2b\sin x\cos x + c{\cos ^2}x,$ then
If $\tan x = \frac{{2b}}{{a - c}}(a \ne c),$
$y = a\,{\cos ^2}x + 2b\,\sin x\cos x + c\,{\sin ^2}x$
and $z = a{\sin ^2}x - 2b\sin x\cos x + c{\cos ^2}x,$ then
$\sin 4\theta $ can be written as
$\sin 4\theta $ can be written as
$\cos 2(\theta + \phi ) - 4\cos (\theta + \phi )\sin \theta \sin \phi + 2{\sin ^2}\phi = $
$\cos 2(\theta + \phi ) - 4\cos (\theta + \phi )\sin \theta \sin \phi + 2{\sin ^2}\phi = $
If $\sin x + \cos x = \frac{1}{5},$ then $\tan 2x$ is
If $\sin x + \cos x = \frac{1}{5},$ then $\tan 2x$ is
The expression,$\frac{{\tan \,\left( {{\textstyle{{3\,\pi } \over 2}}\,\, - \,\,\alpha } \right)\,\,\,\cos \,\left( {{\textstyle{{3\,\pi } \over 2}}\,\, - \,\,\alpha } \right)}}{{\cos \,(2\,\pi \,\, - \,\alpha )}}$ $+ cos \left( {\alpha \,\, - \,\,\frac{\pi }{2}} \right) \,sin (\pi -\alpha ) + cos (\pi +\alpha ) sin \,\left( {\alpha \,\, - \,\,\frac{\pi }{2}} \right)$ when simplified reduces to :
The expression,$\frac{{\tan \,\left( {{\textstyle{{3\,\pi } \over 2}}\,\, - \,\,\alpha } \right)\,\,\,\cos \,\left( {{\textstyle{{3\,\pi } \over 2}}\,\, - \,\,\alpha } \right)}}{{\cos \,(2\,\pi \,\, - \,\alpha )}}$ $+ cos \left( {\alpha \,\, - \,\,\frac{\pi }{2}} \right) \,sin (\pi -\alpha ) + cos (\pi +\alpha ) sin \,\left( {\alpha \,\, - \,\,\frac{\pi }{2}} \right)$ when simplified reduces to :