સાબિત કરો કે, $\frac{\cos 7 x+\cos 5 x}{\sin 7 x-\sin 5 x}=\cot x$
સાબિત કરો કે, $\frac{\cos 7 x+\cos 5 x}{\sin 7 x-\sin 5 x}=\cot x$
we get,
$L.H.S.$ $=\frac{2 \cos \frac{7 x+5 x}{2} \cos \frac{7 x-5 x}{2}}{2 \cos \frac{7 x+5 x}{2} \sin \frac{7 x-5 x}{2}}$
$=\frac{\cos x}{\sin x}=\cot x= R.H.S.$
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- [JEE MAIN 2021]
$\frac{{\tan \,\,\left( {x\,\, - \,\,{\textstyle{\pi \over 2}}} \right)\,\,.\,\,\cos \,\,\left( {{\textstyle{{3\pi } \over 2}}\,\, + \,\,x} \right)\,\, - \,\,{{\sin }^3}\,\left( {{\textstyle{{7\pi } \over 2}}\,\, - \,\,x} \right)}}{{\cos \,\,\left( {x\,\, - \,\,{\textstyle{\pi \over 2}}} \right)\,\,.\,\,\tan \,\,\left( {{\textstyle{{3\pi } \over 2}}\,\, + \,\,x} \right)}}$ =
$\frac{{\tan \,\,\left( {x\,\, - \,\,{\textstyle{\pi \over 2}}} \right)\,\,.\,\,\cos \,\,\left( {{\textstyle{{3\pi } \over 2}}\,\, + \,\,x} \right)\,\, - \,\,{{\sin }^3}\,\left( {{\textstyle{{7\pi } \over 2}}\,\, - \,\,x} \right)}}{{\cos \,\,\left( {x\,\, - \,\,{\textstyle{\pi \over 2}}} \right)\,\,.\,\,\tan \,\,\left( {{\textstyle{{3\pi } \over 2}}\,\, + \,\,x} \right)}}$ =
જો $a\tan \theta = b$, તો $a\cos 2\theta + b\sin 2\theta = $
જો $a\tan \theta = b$, તો $a\cos 2\theta + b\sin 2\theta = $