સાબિત કરો કે : $\cos \left(\frac{\pi}{4}-x\right) \cos \left(\frac{\pi}{4}-y\right)-\sin \left(\frac{\pi}{4}-x\right) \sin \left(\frac{\pi}{4}-y\right)=\sin (x+y)$
$\cos \left(\frac{\pi}{4}-x\right) \cos \left(\frac{\pi}{4}-y\right)-\sin \left(\frac{\pi}{4}-x\right) \sin \left(\frac{\pi}{4}-y\right)$
$=\frac{1}{2}\left[2 \cos \left(\frac{\pi}{4}-x\right) \cos \left(\frac{\pi}{4}-y\right)\right]+\frac{1}{2}\left[-2 \sin \left(\frac{\pi}{4}-x\right) \sin \left(\frac{\pi}{4}-y\right)\right]$
$=\frac{1}{2}\left[\cos \left\{\left(\frac{\pi}{4}-x\right)+\left(\frac{\pi}{4}-y\right)\right\}+\cos \left\{\left(\frac{\pi}{4}-x\right)-\left(\frac{\pi}{4}-y\right)\right\}\right]$
$+\frac{1}{2}\left[\cos \left\{\left(\frac{\pi}{4}-x\right)+\left(\frac{\pi}{4}-y\right)\right\}-\cos \left\{\frac{\pi}{4}-x\right\}-\left(\frac{\pi}{4}-y\right)\right]$
$\left[ \begin{gathered}
\because 2\cos A\cos B = \cos (A + B) + \cos (A - B) \hfill \\
- 2\sin A\sin B = \cos (A + B) - \cos (A - B) \hfill \\
\end{gathered} \right]$
$=2 \times \frac{1}{2}\left[\cos \left\{\left(\frac{\pi}{4}-x\right)+\left(\frac{\pi}{4}-y\right)\right\}\right]$
$=\cos \left[\frac{\pi}{4}-(x+y)\right]$
$=\sin (x+y)$
$= R . H.S$
જો $\sin x + {\sin ^2}x = 1$, તો ${\cos ^{12}}x + 3{\cos ^{10}}x + 3{\cos ^8}x + {\cos ^6}x - 2 =$
જો $\sin \theta + \cos \theta = m$ અને $\sec \theta + {\rm{cosec}}\theta = n$, તો $n(m + 1)(m - 1) = $
$\tan 1^\circ \tan 2^\circ \tan 3^\circ \tan 4^\circ ........\tan 89^\circ = $
સાબિત કરો કે : $(\cos x-\cos y)^{2}+(\sin x-\sin y)^{2}=4 \sin ^{2} \frac{x-y}{2}$
$\cos 1^\circ .\cos 2^\circ .\cos 3^\circ .........\cos 179^\circ = $