Find the value of $4\tan(\theta _1)\tan(\theta _2)$ if $\theta _1$ and $\theta _2$ are the solutions of the given equation Find the value of $4\tan(\theta _1)\tan(\theta _2)$ if $\theta _1$ and $\theta _2$ are the solutions of the given equation:
$$\frac 3{\sec2\theta}+\frac 4{\operatorname {cosec}2\theta}=5$$(1) $0$ (2) $1$ (3) $4$ (4) $16$

Answer (2)

My solution

$$\frac 3{\sec2\theta}+\frac 4{\operatorname {cosec}2\theta}=3\cos2\theta+4\sin2\theta=5\implies \theta _1=\theta_2$$ dividing 5 $$\frac 35\cos2\theta+\frac 45\sin2\theta=1=\sin(37+2\theta)=1=\sin90\implies \theta=\theta_1=\theta_2=\frac{53}{2}$$

so putting the value of $\theta_1$ then $$4\tan(\theta _1)\tan(\theta _2)=4\times \tan^2\theta_1=4\times \tan^2\frac{53}{2}=0.99\approx 1$$(calculated from scientific calculator because I don't know the value of $ \tan(53/2) $

OK, if I were at an exam and this question were there, how would I solve this without knowing the value of $ \tan(53/2) $? What is another method?
For $\tan(\theta)=t$
$$\cos(2\theta)=\frac{1-t^2}{1+t^2}$$

$$\sin(2\theta)=\frac{2t}{1+t^2}$$

Then we have $$\frac 3{\operatorname{sec}2\theta}+\frac 4{\operatorname {cosec}2\theta}=5 \Rightarrow 3(1-t^2)+4(2t)=5(1+t^2)$$  

Now we get a quadratic equation of $t$ , 

$$8t^2+(-8)t+2=0$$

Now suppose that $t_1=\tan(\theta_1) $ and $t_2=\tan(\theta_2)$

Thus by the product of roots we get , $t_1 \cdot t_2=\frac{2}{8}$

So $4t_1 \cdot t_2=1$
HINT:

$$\frac 3{\operatorname{sec}2\theta}+\frac 4{\operatorname {cosec}2\theta}=5$$

Let $$ \frac35 = \sin t , \frac45 = \cos t  $$

$$ \sin (t + 2 \theta) = 1 $$

Can find  two trig angles and take it from there?