સાબિત કરો કે : $\cos \left(\frac{\pi}{4}-x\right) \cos \left(\frac{\pi}{4}-y\right)-\sin \left(\frac{\pi}{4}-x\right) \sin \left(\frac{\pi}{4}-y\right)=\sin (x+y)$
$\cos \left(\frac{\pi}{4}-x\right) \cos \left(\frac{\pi}{4}-y\right)-\sin \left(\frac{\pi}{4}-x\right) \sin \left(\frac{\pi}{4}-y\right)$
$=\frac{1}{2}\left[2 \cos \left(\frac{\pi}{4}-x\right) \cos \left(\frac{\pi}{4}-y\right)\right]+\frac{1}{2}\left[-2 \sin \left(\frac{\pi}{4}-x\right) \sin \left(\frac{\pi}{4}-y\right)\right]$
$=\frac{1}{2}\left[\cos \left\{\left(\frac{\pi}{4}-x\right)+\left(\frac{\pi}{4}-y\right)\right\}+\cos \left\{\left(\frac{\pi}{4}-x\right)-\left(\frac{\pi}{4}-y\right)\right\}\right]$
$+\frac{1}{2}\left[\cos \left\{\left(\frac{\pi}{4}-x\right)+\left(\frac{\pi}{4}-y\right)\right\}-\cos \left\{\frac{\pi}{4}-x\right\}-\left(\frac{\pi}{4}-y\right)\right]$
$\left[ \begin{gathered}
\because 2\cos A\cos B = \cos (A + B) + \cos (A - B) \hfill \\
- 2\sin A\sin B = \cos (A + B) - \cos (A - B) \hfill \\
\end{gathered} \right]$
$=2 \times \frac{1}{2}\left[\cos \left\{\left(\frac{\pi}{4}-x\right)+\left(\frac{\pi}{4}-y\right)\right\}\right]$
$=\cos \left[\frac{\pi}{4}-(x+y)\right]$
$=\sin (x+y)$
$= R . H.S$
$6$ રેડિયનને અંશ માપમાં ફેરવો.
કિંમત શોધો : $\sin 75^{\circ}$
જો $\frac{{3\pi }}{4} < \alpha < \pi ,$ તો $\sqrt {{\rm{cose}}{{\rm{c}}^2}\alpha + 2\cot \alpha } = . . .$
જો $A$ એ બીજા ચરણમાં હોય અને $3\tan A + 4 = 0,$ તો $2\cot A - 5\cos A + \sin A$ ની કિમત મેળવો.
${\sin ^2}{5^o} + {\sin ^2}{10^o} + {\sin ^2}{15^o} + ... + $ ${\sin ^2}{85^o} + {\sin ^2}{90^o} = . ..$