$\left( {1 + \cos \frac{\pi }{9}} \right)\left( {1 + \cos \frac{{3\pi }}{9}} \right)\left( {1 + \cos \frac{{5\pi }}{9}} \right)\left( {1 + \cos \frac{{7\pi }}{9}} \right)$ની કિમત ............ થાય
$\left( {1 + \cos \frac{\pi }{9}} \right)\left( {1 + \cos \frac{{3\pi }}{9}} \right)\left( {1 + \cos \frac{{5\pi }}{9}} \right)\left( {1 + \cos \frac{{7\pi }}{9}} \right)$ની કિમત ............ થાય
- A
$\frac{9}{{16}}\,$
- B
$\frac{10}{{16}}\,$
- C
$\frac{12}{{16}}\,$
- D
$\frac{5}{{16}}\,$
Similar Questions
$\frac{{\tan \,\,\left( {x\,\, - \,\,{\textstyle{\pi \over 2}}} \right)\,\,.\,\,\cos \,\,\left( {{\textstyle{{3\pi } \over 2}}\,\, + \,\,x} \right)\,\, - \,\,{{\sin }^3}\,\left( {{\textstyle{{7\pi } \over 2}}\,\, - \,\,x} \right)}}{{\cos \,\,\left( {x\,\, - \,\,{\textstyle{\pi \over 2}}} \right)\,\,.\,\,\tan \,\,\left( {{\textstyle{{3\pi } \over 2}}\,\, + \,\,x} \right)}}$ =
$\frac{{\tan \,\,\left( {x\,\, - \,\,{\textstyle{\pi \over 2}}} \right)\,\,.\,\,\cos \,\,\left( {{\textstyle{{3\pi } \over 2}}\,\, + \,\,x} \right)\,\, - \,\,{{\sin }^3}\,\left( {{\textstyle{{7\pi } \over 2}}\,\, - \,\,x} \right)}}{{\cos \,\,\left( {x\,\, - \,\,{\textstyle{\pi \over 2}}} \right)\,\,.\,\,\tan \,\,\left( {{\textstyle{{3\pi } \over 2}}\,\, + \,\,x} \right)}}$ =
$\frac{{\sin 3A - \cos \left( {\frac{\pi }{2} - A} \right)}}{{\cos A + \cos (\pi + 3A)}} = $
$\frac{{\sin 3A - \cos \left( {\frac{\pi }{2} - A} \right)}}{{\cos A + \cos (\pi + 3A)}} = $
જો $cosA + cosB = cosC,\ sinA + sinB = sinC$ હોય તો સમીકરણ $\frac{{\sin \left( {A + B} \right)}}{{\sin 2C}}$ =
જો $cosA + cosB = cosC,\ sinA + sinB = sinC$ હોય તો સમીકરણ $\frac{{\sin \left( {A + B} \right)}}{{\sin 2C}}$ =
$\left( {\frac{{\sin 2A}}{{1 + \cos 2A}}} \right)\,\left( {\frac{{\cos A}}{{1 + \cos A}}} \right)= $
$\left( {\frac{{\sin 2A}}{{1 + \cos 2A}}} \right)\,\left( {\frac{{\cos A}}{{1 + \cos A}}} \right)= $
સાબિત કરો કે : $\frac{\sin x-\sin 3 x}{\sin ^{2} x-\cos ^{2} x}=2 \sin x$
સાબિત કરો કે : $\frac{\sin x-\sin 3 x}{\sin ^{2} x-\cos ^{2} x}=2 \sin x$