Prove that:
$2 \sin ^{2} \frac{\pi}{6}+\cos ec ^{2} \frac{7 \pi}{6} \cos ^{2} \frac{\pi}{3}=\frac{3}{2}$
$L.H.S.$ $=2 \sin ^{2} \frac{\pi}{6}+\cos ec ^{2}\, \frac{7 \pi}{6} \,\cos ^{2} \,\frac{\pi}{3}$
$=2\left(\frac{1}{2}\right)^{2}+\cos ec ^{2}\left(\pi+\frac{\pi}{6}\right)\left(\frac{1}{2}\right)^{2}$
$=2 \times \frac{1}{4}+\left(-\cos ec \,\frac{\pi}{6}\right)^{2}\left(\frac{1}{4}\right)$
$=\frac{1}{2}+(-2)^{2}\left(\frac{1}{4}\right)$
$=\frac{1}{2}+\frac{4}{4}=\frac{1}{2}+1=\frac{3}{2}$
$=R .H.S.$
Find the degree measures corresponding to the following radian measures ( Use $\pi=\frac{22}{7}$ ).
$\frac{7 \pi}{6}$
Prove that :
$\cot ^{2} \frac{\pi}{6}+\cos ec \,\frac{5 \pi}{6}+3 \tan ^{2}\, \frac{\pi}{6}=6$
The value of $\cos A - \sin A$ when $A = \frac{{5\pi }}{4},$ is
$\frac{{\sin \theta }}{{1 - \cot \theta }} + \frac{{\cos \theta }}{{1 - \tan \theta }} = $
Let the function $:(0, \pi) \rightarrow R$ be defined by
$f (\theta)=(\sin \theta+\cos \theta)^2+(\sin \theta-\cos \theta)^4$
Suppose the function $f$ has a local minimum at $\theta$ precisely when $\theta \in\left\{\lambda_1 \pi, \ldots, \lambda_{ T } \pi\right\}$, where $0<\lambda_1<\cdots<\lambda_r<1$. Then the value of $\lambda_1+\cdots+\lambda_r$ is. . . . .