Question

Prove that $\tan 7\alpha -\tan 5\alpha -\tan 2\alpha =\tan 7\alpha .\tan 5\alpha .\tan 2\alpha$

Answer 1

We solve this problem by using a condition such that the given angles will satisfy and then apply the tangent trigonometric function on both sides to get the required answer. Here, the condition that satisfies the given three angles is
$7\alpha =5\alpha +2\alpha$
By applying the tangent trigonometric function on both sides we use the following formula of composite angles to get the answer.
$\tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A.\tan B}$
Also, we can take
$\tan \left( A-B \right)=\dfrac{\tan A-\tan B}{1+\tan A.\tan B}$

Complete step by step answer:
We are asked to prove the result
$\tan 7\alpha -\tan 5\alpha -\tan 2\alpha =\tan 7\alpha .\tan 5\alpha .\tan 2\alpha$
Here, we can see that there are three angles $7\alpha ,5\alpha ,2\alpha$.
Here, we can find the condition between the three angles as follows
$\Rightarrow 7\alpha =5\alpha +2\alpha$
Now, by applying the tangent trigonometric function on both sides we get
$\Rightarrow \tan 7\alpha =\tan \left( 5\alpha +2\alpha \right)$
We know that the formula of composite angles as
$\tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A.\tan B}$
Now, by applying the above formula we get
$\Rightarrow \tan 7\alpha =\dfrac{\tan 5\alpha +\tan 2\alpha }{1-\tan 5\alpha .\tan 2\alpha }$
Now, by cross multiplying we get

& \Rightarrow \tan 7\alpha \left( 1-\tan 5\alpha .\tan 2\alpha \right)=\tan 5\alpha +\tan 2\alpha \\\ & \Rightarrow \tan 7\alpha -\tan 7\alpha .\tan 5\alpha .\tan 2\alpha =\tan 5\alpha +\tan 2\alpha \\\ \end{aligned}$$ Now, by rearranging the terms from both sides we get $$\Rightarrow \tan 7\alpha -\tan 5\alpha -\tan 2\alpha =\tan 7\alpha .\tan 5\alpha .\tan 2\alpha $$ Therefore, we can say that the required result has been proved. **Note:** We can solve this problem by a different method. Here, we can see that there are three angles $$7\alpha ,5\alpha ,2\alpha $$. Here, we can find the condition between the three angles as follows $$\Rightarrow 7\alpha -5\alpha =2\alpha $$ Now, by applying the tangent trigonometric function on both sides we get $$\Rightarrow \tan 2\alpha =\tan \left( 7\alpha -5\alpha \right)$$ We know that the formula of composite angles as $$\tan \left( A-B \right)=\dfrac{\tan A-\tan B}{1+\tan A.\tan B}$$ Now, by applying the above formula we get $$\Rightarrow \tan 2\alpha =\dfrac{\tan 7\alpha -\tan 5\alpha }{1+\tan 5\alpha .\tan 7\alpha }$$ Now, by cross multiplying we get $$\begin{aligned} & \Rightarrow \tan 2\alpha \left( 1+\tan 5\alpha .\tan 7\alpha \right)=\tan 7\alpha -\tan 5\alpha \\\ & \Rightarrow \tan 2\alpha +\tan 7\alpha .\tan 5\alpha .\tan 2\alpha =\tan 7\alpha -\tan 5\alpha \\\ \end{aligned}$$ Now, by rearranging the terms from both sides we get $$\Rightarrow \tan 7\alpha -\tan 5\alpha -\tan 2\alpha =\tan 7\alpha .\tan 5\alpha .\tan 2\alpha $$ Therefore, we can say that the required result has been proved. Here, we can use the other condition of three angles as $$\Rightarrow 7\alpha -2\alpha =5\alpha $$ In this process also we get the same result as $$\Rightarrow \tan 7\alpha -\tan 5\alpha -\tan 2\alpha =\tan 7\alpha .\tan 5\alpha .\tan 2\alpha $$ Therefore, we can say that the required result has been proved.