$$6 \cos \theta - 7 \sin \theta = 0 \Rightarrow (7 \cos 2\theta + 6 \sin 2\theta)^2 =$$

 we have given that 

   $$6 \cos \theta - 7 \sin \theta = 0 $$              equation 1  and

   $$(7 \cos 2\theta + 6 \sin 2\theta)^2 =$$        equation 2 

then solving equation 1 we get 

$$6 \cos \theta - 7 \sin \theta = 0 $$

$$ \tan \theta$$ = 6\7 

now solving equation 2 

$$(7 \cos 2\theta + 6 \sin 2\theta)^2 =$$    here we know that 

$$ \cos 2\theta$$ = $$\cos\theta^{2}$$ - $$\sin\theta^{2}$$  and   $$ \sin 2\theta$$ = 2$$\sin\theta$$ $$\cos\theta$$

put the value in equation 2 we get 

=  (7($$\cos\theta^{2}$$ - $$\sin\theta^{2}$$) + 6 ( 2$$ \sin \theta$$ $$\cos\theta$$))^2  

on solving ge get 

= 7\6$$\times$$ $$tan\theta$$ $$7^{2}$$

= 49 answer