Question

How do you write the expression $\dfrac{\tan 2x+\tan x}{1-\tan 2x\tan x}$ as the sine, cosine or the tangent of an angle?

Answer 1

In order to do this question, you need to know the tan(A + B) formula. By using this you can simplify the above equation. $\tan \left( a+b \right)=\dfrac{\tan a+\tan b}{1-\tan a\tan b}$. Here a is 2x and b is x. Therefore, you can simplify it. Next you need to know the formula of tan in terms of sine and cos that is $\tan x=\dfrac{\sin x}{\cos x}$. Therefore, you can write tan in terms of sine and cosine.

Complete step-by-step solution:
The first step in order to solve this question is to first simplify the $\dfrac{\tan 2x+\tan x}{1-\tan 2x\tan x}$ . Here we use the formula of tan (a + b) which is given by $\tan \left( a+b \right)=\dfrac{\tan a+\tan b}{1-\tan a\tan b}$. Now, we have to take a is 2x and b is x. Therefore, we can simplify the expression as
$\Rightarrow \tan \left( 2x+x \right)=\dfrac{\tan 2x+\tan x}{1-\tan 2x\tan x}$
$\Rightarrow \tan \left( 3x \right)=\dfrac{\tan 2x+\tan x}{1-\tan 2x\tan x}$
Therefore, in terms of tan, we can write the above expression as tan(3x).
Next step we have to do is that we have to write tan in terms of sin and cos. For this, we use the formula $\tan x=\dfrac{\sin x}{\cos x}$. Therefore we get the answer as
$\Rightarrow \tan \left( 3x \right)=\dfrac{\sin 3x}{\cos 3x}$.
Therefore, we get the final answer of the question how do you write the expression $\dfrac{\tan 2x+\tan x}{1-\tan 2x\tan x}$ as the sine, cosine or the tangent of an angle as $tan\left( 3x \right)=\dfrac{\sin 3x}{\cos 3x}$.

Note: In order to do this question, you need to know the trigonometric identities and the sum of angles for tan and cosine and sine. Otherwise, you cannot solve this problem. Also, you have to be careful while using the formulas and while substitutions.