Concept of Trigonometric identities The value of the expression 
$$\dfrac{\sin x}{ \cos 3x} + \dfrac{\sin 3x}{ \cos 9x} + \dfrac{\sin 9x}{ \cos 27x}$$ in terms of $\tan x$ is  
My Approach
If I take L.C.M of this as $\cos 3 \cos 9x \cos 27x$ and respectively multiply the numerator then it is getting very lengthy. Even if I will use the identity of $\sin 3x$ then also I am not getting appropriate answer. Please suggest some nice and short way of doing this question. 

$$ {\sin x \over \cos 3x}$$ 
$$= {\sin x \over \cos x(1 - 4\sin^2 x)}$$
$$= {\tan x \over (1 - 4\sin^2 x)}$$

Now you can get $$\sin^2 x = {\tan^2 x \over 1 + \tan^2 x}$$ from $$ \cot^2 x + 1= { 1\over \sin^2 x}$$

$$\therefore {\sin x \over \cos 3x} = {\tan x \over (1 - 4\sin^2 x)} = {\tan x (1 + \tan^2 x) \over (1 - 3\tan^2 x)}$$
