${(\cos \alpha + \cos \beta )^2} + {(\sin \alpha + \sin \beta )^2} = $
${(\cos \alpha + \cos \beta )^2} + {(\sin \alpha + \sin \beta )^2} = $
- A
$4{\cos ^2}\frac{{\alpha - \beta }}{2}$
- B
$4{\sin ^2}\frac{{\alpha - \beta }}{2}$
- C
$4{\cos ^2}\frac{{\alpha + \beta }}{2}$
- D
$4{\sin ^2}\frac{{\alpha + \beta }}{2}$
Similar Questions
$ \cos ^{3}\left(\frac{\pi}{8}\right) \cdot \cos \left(\frac{3 \pi}{8}\right)+\sin ^{3}\left(\frac{\pi}{8}\right) \cdot \sin \left(\frac{3 \pi}{8}\right)$ ની કિમંત મેળવો.
$ \cos ^{3}\left(\frac{\pi}{8}\right) \cdot \cos \left(\frac{3 \pi}{8}\right)+\sin ^{3}\left(\frac{\pi}{8}\right) \cdot \sin \left(\frac{3 \pi}{8}\right)$ ની કિમંત મેળવો.
- [JEE MAIN 2020]
જો $x = sec\, \phi - tan\, \phi$ & $y = cosec\, \phi + cot\, \phi$ હોય તો,
જો $x = sec\, \phi - tan\, \phi$ & $y = cosec\, \phi + cot\, \phi$ હોય તો,
$\tan {3^o} + 2\tan {6^o} + 4\tan {12^o} + 8\cot {24^o} = \cot {\theta ^o}$ થાય તો
$\tan {3^o} + 2\tan {6^o} + 4\tan {12^o} + 8\cot {24^o} = \cot {\theta ^o}$ થાય તો
$\frac{{4\sin {9^o}\sin {{21}^o}\sin {{39}^o}\sin {{51}^o}\sin {{69}^o}\sin {{81}^o}}}{{\sin {{54}^o}}}$ =
$\frac{{4\sin {9^o}\sin {{21}^o}\sin {{39}^o}\sin {{51}^o}\sin {{69}^o}\sin {{81}^o}}}{{\sin {{54}^o}}}$ =
સાબિત કરો કે : $\frac{\sin x-\sin 3 x}{\sin ^{2} x-\cos ^{2} x}=2 \sin x$
સાબિત કરો કે : $\frac{\sin x-\sin 3 x}{\sin ^{2} x-\cos ^{2} x}=2 \sin x$