<s>[INST] Find Value of $(3\tan^2\alpha \;+ 4\tan^2\beta)$ Let $\vec V_1$ and $\vec V_2$ are two vectors such that $\vec V_1= 2(\sin\alpha+\cos\alpha) \hat i+\hat j$ and $\vec V_2=\sin\beta \; \hat i +\cos\beta \; \hat j$, where $\alpha$ and $\beta$ satisfy the relation $2 (\sin \alpha \; + \cos\alpha)\sin\beta=3-\cos\beta, $ Find Value of $(3\tan^2\alpha \;+ 4\tan^2\beta)$
My Approach:
I took Dot Product of $\vec V_1$ and $\vec V_2$ $\;$ and I obtained $\vec V_1.\vec V_2=3 $ using the given Relation.
Solving Given Relation too did not lead me anywhere
I am stuck now.
How to processed further?
 [/INST] Trick question! Doesn't really have anything to do with vectors.
Note that the maximum value of $a \sin \theta+b \cos \theta$ is $\sqrt {a^2+b^2}$, which occurs when $\tan \theta =\frac ab$.
Here, we've been given that:
$$2(\sin \alpha+ \cos \alpha)\sin \beta+\cos \beta=3$$
Now, maximum value of LHS is $$\sqrt{4(\sin \alpha+\cos \alpha)^2+1}=\sqrt {5+4\sin 2\alpha}$$
Notice that $\sqrt {5+4\sin 2\alpha}\leq 3$, reaching maximum value at $\sin 2\alpha=1$.
Thus, maximum value of LHS, is $3$, which is reached when $\sin 2\alpha=1$ and $\tan \beta=2(\sin \alpha+\cos \alpha)$.
Now, $\sin 2\alpha=1 \implies \tan^2 \alpha=1$, and $\tan^2 \beta=4(1+\sin 2\alpha)=8$.
Thus, $3\tan^2 \alpha +4\tan^2 \beta=3\cdot 1+4\cdot 8=35$.
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