Odisha Board Solutions for Chapter: Trigonometric Functions, Exercise 2: Exercises- 4(b)

If $\theta$ lies in third quadrant, than $\cos \frac{\theta }{2}+\sin \frac{\theta }{2}$ is positive.

$2\sin 105°.\sin 15°=\frac{1}{2}$

If $\cos A=\frac{1}{2},\cos B=1$ then $\tan \frac{A+B}{2}.\tan \frac{A-B}{2}=1$, given $A,B$ lies in first quadrant

$\cos 15°\cos 7\frac{1}{2}°·\sin 7\frac{1}{2}°=1$.

$\sin 20°\left(3-4{\cos }^{2}70°\right)=\frac{\sqrt{3}}{2}$.

$\sqrt{3}\left(3\tan 10°-{\tan }^{3}10°\right)=\left(1-3{\tan }^{2}10°\right)$

$\frac{2\tan 7{\displaystyle \frac{1}{2}}°\left(1-{\tan }^{2}7\frac{1}{2}°\right)}{\left(1+{\tan }^{2}7\frac{1}{2}°\right)}=1$

The minimum value of $\sin \theta .\cos \theta$ is ${\left(-1\right)}^2$.