સાબિત કરો કે : $\frac{(\sin 7 x+\sin 5 x)+(\sin 9 x+\sin 3 x)}{(\cos 7 x+\cos 5 x)+(\cos 9 x+\cos 3 x)}=\tan 6 x$
It is known that
$\sin A+\sin B=2 \sin \left(\frac{A+B}{2}\right) \cdot \cos \left(\frac{A-B}{2}\right), \cos A+\cos B=2 \cos \left(\frac{A+B}{2}\right) \cdot \cos \left(\frac{A-B}{2}\right)$
$L.H.S.$ $=\frac{(\sin 7 x+\sin 5 x)+(\sin 9 x+\sin 3 x)}{(\cos 7 x+\cos 5 x)+(\cos 9 x+\cos 3 x)}$
$=\frac{\left[2 \sin \left(\frac{7 x+5 x}{2}\right) \cdot \cos \left(\frac{7 x-5 x}{2}\right)\right]+\left[2 \sin \left(\frac{9 x+3 x}{2}\right) \cdot \cos \left(\frac{9 x-3 x}{2}\right)\right]}{\left[2 \cos \left(\frac{7 x+5 x}{2}\right) \cdot \cos \left(\frac{7 x-5 x}{2}\right)\right]+\left[2 \cos \left(\frac{9 x+3 x}{2}\right) \cdot \cos \left(\frac{9 x-3 x}{2}\right)\right]}$
$=\frac{[2 \sin 6 x \cdot \cos x]+[2 \sin 6 x \cdot \cos 3 x]}{[2 \cos 6 x \cdot \cos x]+[2 \cos 6 x \cdot \cos 6 x]}$
$=\frac{2 \sin 6 x[\cos x+\cos 3 x]}{2 \cos 6 x[\cos x+\cos 3 x]}$
$=\tan 6 x$
$= R . H.S.$
જો $\theta $ અને $\phi $ એ પ્રથમ ચરણમાં આવેલ છે કે જેથી $\tan \theta = 1/7$ અને $\sin \phi = 1/\sqrt {10} $.તો
$\tan \frac{13 \pi}{12}$ નું મૂલ્ય શોધો.
જો $\cos x + {\cos ^2}x = 1,$ તો ${\sin ^2}x + {\sin ^4}x =$
સાબિત કરો કે : $\sin 3 x+\sin 2 x-\sin x=4 \sin x \cos \frac{x}{2} \cos \frac{3 x}{2}$
જો $\sin x + {\rm{cosec}}\,x = 2,$ તો $sin^n x + cosec^n x = .. . .$