સાબિત કરો કે : $\tan 4 x=\frac{4 \tan x\left(1-\tan ^{2} x\right)}{1-6 \tan ^{2} x+\tan ^{4} x}$
It is known that $\tan 2 A=\frac{2 \tan A}{1-\tan ^{2} A}$
$\therefore$ $L.H.S.$ $=\tan 4 x=\tan 2(2 x)$
$=\frac{2 \tan 2 x}{1-\tan ^{2}(2 x)}$
$=\frac{2\left(\frac{2 \tan x}{1-\tan ^{2} x}\right)}{1-\left(\frac{2 \tan x}{1-\tan ^{2} x}\right)^{2}}$
$=\frac{\left(\frac{4 \tan x}{1-\tan ^{2} x}\right)}{\left[1-\frac{4 \tan ^{2} x}{\left(1-\tan ^{2} x\right)^{2}}\right]}$
$=\frac{\left(\frac{4 \tan x}{1-\tan ^{2} x}\right)}{\left[\frac{\left(1-\tan ^{2} x\right)^{2}-4 \tan ^{2} x}{\left(1-\tan ^{2} x\right)^{2}}\right]}$
$=\frac{4 \tan x\left(1-\tan ^{2} x\right)}{\left(1-\tan ^{2} x\right)^{2}-4 \tan ^{2} x}$
$=\frac{4 \tan x\left(1-\tan ^{2} x\right)}{1+\tan ^{4} x-2 \tan ^{2} x-4 \tan ^{2} x}$
$=\frac{4 \tan x\left(1-\tan ^{2} x\right)}{1-6 \tan ^{2} x+\tan ^{4} x}= R . H.S.$
જો $x\, sin \theta = y\, sin \, \left( {\theta \,\, + \,\,\frac{{2\,\pi }}{3}} \right) = z\, sin \, \left( {\theta \,\, + \,\,\frac{{4\,\pi }}{3}} \right)$ હોય તો
$\frac{{\tan \,\,\left( {x\,\, - \,\,{\textstyle{\pi \over 2}}} \right)\,\,.\,\,\cos \,\,\left( {{\textstyle{{3\pi } \over 2}}\,\, + \,\,x} \right)\,\, - \,\,{{\sin }^3}\,\left( {{\textstyle{{7\pi } \over 2}}\,\, - \,\,x} \right)}}{{\cos \,\,\left( {x\,\, - \,\,{\textstyle{\pi \over 2}}} \right)\,\,.\,\,\tan \,\,\left( {{\textstyle{{3\pi } \over 2}}\,\, + \,\,x} \right)}}$ =
જો $3\cos \theta + 4\sin \theta = 5$ અને $3\sin \theta - 4\cos \theta $ =
$\sin 4\theta $ ને . . . . સ્વરૂપે પણ લખી શકાય.
જો $x = \sin {130^o}\,\cos {80^o},\,\,y = \sin \,{80^o}\,\cos \,{130^o},\,\,z = 1 + xy,$ તો આપેલ પૈકી ક્યૂ સત્ય છે.