સાબિત કરો કે : $\cos \left(\frac{3 \pi}{4}+x\right)-\cos \left(\frac{3 \pi}{4}-x\right)=-\sqrt{2} \sin x$
It is known that $\cos A-\cos B=-2 \sin \left(\frac{A+B}{2}\right) \cdot \sin \left(\frac{A-B}{2}\right)$
$\therefore$ $L.H.S.$ $=\cos \left(\frac{3 \pi}{4}+x\right)-\cos \left(\frac{3 \pi}{4}-x\right)$
$=-2 \sin \left\{\frac{\left(\frac{3 \pi}{4}+x\right)+\left(\left(\frac{3 \pi}{4}-x\right)\right)}{2}\right\} \cdot \sin \left\{\frac{\left(\frac{3 \pi}{4}+x\right)-\left(\frac{3 \pi}{4}-x\right)}{2}\right\}$
$=-2 \sin \left(\frac{3 \pi}{4}\right) \sin x$
$=-2 \sin \left(\pi-\frac{\pi}{4}\right) \sin x$
$=-2 \sin \frac{\pi}{4} \sin x$
$=-2 \times \frac{1}{\sqrt{2}} \times \sin x$
$=-\sqrt{2} \sin x$
$= R . H.S.$
કિંમત શોધો : $\tan 15^{\circ}$
$6({\sin ^6}\theta + {\cos ^6}\theta ) - 9({\sin ^4}\theta + {\cos ^4}\theta ) + 4 = . . . $
જો $A = 130^\circ $ અને $x = \sin A + \cos A,$ તો
જો $\tan \theta = \frac{{ - 4}}{3},$ તો $\sin \theta = $
જો $\sin \theta + {\rm{cosec}}\theta = {\rm{2}}$, તો ${\sin ^2}\theta + {\rm{cose}}{{\rm{c}}^{\rm{2}}}\theta = $