Given $$3$$ $$\sin \,P + 4\cos Q = 6$$ $$\,\,\,\,\,\,\,\,...\left( i \right)$$

$$4\sin Q + 3\cos P = 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$$

Squaring and adding $$(i)$$ & $$(ii)$$ we get

$$9\,{\sin ^2}P + 16{\cos ^2}Q + 24\sin P\cos Q$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + 16\,{\sin ^2}Q + 9{\cos ^2}P + 24\sin Q\cos P$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$ = 36 + 1 = 37$$

$$ \Rightarrow 9\left( {{{\sin }^2}p + {{\cos }^2}P} \right) + 16\left( {{{\sin }^2}Q + {{\cos }^2}q} \right)$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + 24\left( {\sin P\cos Q + \cos P\sin Q} \right) = 37$$

$$ \Rightarrow 9 + 16 + 24\sin \left( {P + Q} \right) = 37$$

[ As $${\sin ^2}\theta + {\cos ^2}\theta = 1$$ and

$$\sin A\cos B + \cos A\sin B$$ $$ = \sin \left( {A + B} \right)$$ ]

$$ \Rightarrow \sin \left( {P + Q} \right) = {1 \over 2}$$

$$ \Rightarrow P + Q = {\pi \over 6}$$ or $${{5\pi } \over 6}$$

$$ \Rightarrow R = {{5\pi } \over 6}$$ or $${\pi \over 6}$$

(as $$P + Q + R = \pi $$ )

If $$R = {{5\pi } \over 6}$$ then $$0 < P,Q < {\pi \over 6}$$

$$ \Rightarrow \cos Q < 1$$ and $$\sin P < {1 \over 2}$$

$$ \Rightarrow 3\sin P + 4\cos Q < {{11} \over 2}$$ which is not true.

So $$R = {\pi \over 6}$$

$$A = {\sin ^2}x + {\cos ^4}x$$

$$ = {\sin ^2}x + {\cos ^2}x\left( {1 - {{\sin }^2}x} \right)$$

$$ = {\sin ^2}x + {\cos ^2}x - {1 \over 4}{\left( {2\sin x.\cos x} \right)^2}$$

$$ = 1 - {1 \over 4}{\sin ^2}\left( {2x} \right)$$

Now $$0 \le {\sin ^2}\left( {2x} \right) \le 1$$

$$ \Rightarrow 0 \ge - {1 \over 4}{\sin ^2}\left( {2x} \right) \ge - {1 \over 4}$$

$$ \Rightarrow 1 \ge 1 - {1 \over 4}{\sin ^2}\left( {2x} \right) \ge 1 - {1 \over 4}$$

$$ \Rightarrow 1 \ge A \ge {3 \over 4}$$

$$\cos \left( {\alpha + \beta } \right) = {4 \over 5} \Rightarrow \tan \left( {\alpha + \beta } \right) = {3 \over 4}$$

$$\sin \left( {\alpha - \beta } \right) = {5 \over {13}} \Rightarrow \tan \left( {\alpha - \beta } \right) = {5 \over {12}}$$

$$\tan 2\alpha = \tan \left[ {\left( {\alpha + \beta } \right) + \left( {\alpha - \beta } \right)} \right]$$

$$ = {{{3 \over 4} + {5 \over {12}}} \over {1 - {3 \over 4}.{5 \over {12}}}} = {{56} \over {33}}$$