Double Angle Formulas

We have, 

$$\operatorname{cosec}10°-\sqrt{3}\sec10°\ =\dfrac{1}{\sin10^{\circ\ }}-\dfrac{\sqrt{\ 3}}{\cos10^{\circ\ }}$$

$$=\dfrac{\cos10^{\circ\ }-\sqrt{\ 3}\sin10^{\circ\ }}{\sin10^{\circ\ }\cos10^{\circ\ }}$$

$$=2\times\ 2\times\ \dfrac{\frac{1}{2}\left(\cos10^{\circ\ }-\sqrt{\ 3}\sin10^{\circ\ }\right)}{2\times\ \sin10^{\circ\ }\cos10^{\circ\ }}$$

$$=4\times\ \dfrac{\frac{1}{2}\cos10^{\circ\ }-\dfrac{\sqrt{\ 3}}{2}\sin10^{\circ\ }}{2\times\sin10^{\circ\ }\cos10^{\circ\ }}$$

$$=4\times\ \dfrac{\sin30^{\circ\ }\cos10^{\circ\ }-\cos30^{\circ\ }\sin10^{\circ\ }}{2\sin10^{\circ\ }\cos10^{\circ\ }}$$

$$=4\times\ \dfrac{\sin\left(30-10\right)^{\circ\ }}{\sin\left(2\times\ 10\right)^{\circ\ }}=4\times\ \dfrac{\sin20^{\circ\ }}{\sin20^{\circ\ }}=4$$

The correct answer is Option A (4).

$$\sqrt{3}\cos2\theta+8\cos\theta+3\sqrt{3}=0$$

We know that $$\cos2\theta=2cos^2\theta-1$$

$$\sqrt{3}\left(2\cos^2\theta-1\right)+8\cos\theta+3\sqrt{3}=0$$ 

$$2\sqrt{3}\cos^2\theta-\sqrt{3}+8\cos\theta+3\sqrt{3}=0$$

$$2\sqrt{3}\cos^2\theta+8\cos\theta+2\sqrt{3}=0$$

$$2\sqrt{3}\cos^2\theta+6\cos\theta+2\cos\theta+2\sqrt{3}=0$$

$$2\sqrt{3}\cos\theta(\cos\theta+\sqrt{3})+2(\cos\theta+\sqrt{3})=0$$

$$(2\sqrt{3}\cos\theta+2)(\cos\theta+\sqrt{3})=0$$

Thus, there are two solutions possible -> $$\cos\theta=-\dfrac{1}{\sqrt{3}}$$ and $$\cos\theta=-\sqrt{3}$$

Since, the value of $$\cos\theta\epsilon[-1,1]$$, therefore, $$\cos\theta=-\sqrt{3}$$ is never possible. 

Now, $$\cos\theta$$ is negative in the second and the third quadrant, and in each cycle of $$\pi$$, $$\cos\theta$$ takes all values in $$[-1,1]$$. Thus, in every cycle of $$\pi$$, there will be one solution for $$\cos\theta=-\dfrac{1}{\sqrt{3}}$$.

Thus, from $$[-3\pi,2\pi]$$, there will be 5 solutions for this equation.

We have, 

$$f(\theta)=4\left(\sin^4\left(\frac{7\pi}{2}-\theta\right)+\sin^4(11\pi+\theta)\right)-2\left(\sin^6\left(\frac{3\pi}{2}-\theta\right)+\sin^6(9\pi-\theta)\right)$$

or, $$f(\theta)=4\left(\cos^4\theta+\sin^4\theta\right)-2\left(\cos^6\theta+\sin^6\theta\right)$$

or, $$f(\theta)=4\left(\cos^4\theta+\sin^4\theta\right)-2\left(\cos^2\theta+\sin^2\theta\right)\left(\cos^4\theta+\cos^2\theta\sin^2\theta+\sin^4\theta\right)$$

or, $$f(\theta)=4\left(\cos^4\theta+\sin^4\theta\right)-2\left(\cos^4\theta+\cos^2\theta\ \sin^2\theta+\sin^4\theta\right)$$

or, $$f(\theta)=4\cos^4\theta+4\sin^4\theta-2\cos^4\theta+2\cos^2\theta\ \sin^2\theta-2\sin^4\theta$$

or, $$f(\theta)=2\cos^4\theta+2\sin^4\theta+2\cos^2\theta\ \sin^2\theta=2\left(\cos^4\theta+\sin^4\theta+\cos^2\theta\ \sin^2\theta\right)$$

or, $$f(\theta)=2\left(\cos^4\theta+\sin^4\theta+2\cos^2\theta\sin^2\theta-\cos^2\theta\sin^2\theta\right)=2\left(1-\cos^2\theta\sin^2\theta\right)$$

or, $$f(\theta)=2\left(1-\cos^2\theta\ \sin^2\theta\right)=2-2\cos^2\theta\ \sin^2\theta=2-\dfrac{\sin^22\theta}{2}\ $$

This expression has maximum and minimum values as $$\alpha$$ and $$\beta$$ respectively.

The maximum and minimum value for $$\sin^22\theta$$ is 1 and 0 respectively for $$\theta\in R\ $$. 

Hence, 

$$f(\theta)_{\min}=2-\dfrac{1}{2}=\dfrac{3}{2}\ =\beta\ $$

$$f(\theta)_{\max}=2-\dfrac{0}{2}=2\ =\alpha\ $$

So, $$\alpha+2\beta\ =2+2\times\dfrac{3}{2}=2+3=5$$