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Hint: Try to use the identity $\tan \left( A+B \right)=\dfrac{\tan A+tanB}{1-\tan A\tan B}$. Then substitute ‘B’ as ‘2A’ and do further calculations to get the result.

Complete step-by-step answer:

For proving what is the asked in the question we have to use the identity

$\tan \left( A+B \right)=\dfrac{\tan A+tanB}{1-\tan A\tan B}$

Where A and B are any 2 angles.

First of all we have to prove the identity by using the facts.

\[\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B,cos\left( A+B \right)=\cos A\cos B-\sin A\sin B\]

Consider the identity,$\tan \left( A+B \right)=\dfrac{\sin (A+B)}{\cos (A+B)}$

Now we can write it as,

$\tan \left( A+B \right)=\dfrac{\text{sinA cosB+cosA sinB}}{\text{cosA cosB-sinA sinB}}$

Now dividing by $\cos A\cos B$ in numerator and denominator we get,

$\tan \left( A+B \right)=\dfrac{\dfrac{\text{sinA cosB}}{\text{cosA cosB}}\text{+}\dfrac{\text{cosA sinB}}{\text{cosA cosB}}}{\dfrac{\text{cosA cosB}}{\text{cosA cosB}}\text{ - }\dfrac{\text{sinA sinB}}{\text{cosA cosB}}}$

Cancelling the like terms, we get

$\tan \left( A+B \right)=\dfrac{\operatorname{tanA}+tanB}{\text{1-tanA tanB}}$

Now we can use the identity by taking ‘B’ as ‘2A’

Here by substituting B as 2A we get,

$\tan \left( A+2A \right)=\dfrac{\operatorname{tanA}+tan2A}{\text{1-tanA tan2A}}$

On further simplification we get,

$\tan \left( 3A \right)=\dfrac{\operatorname{tanA}+tan2A}{\text{1-tan2A tanA}}$

Now on cross multiplication we get,

$\tan \left( 3A \right)\left\{ 1-\tan A\tan 2 \right.\left. A \right\}=\tan A+\tan 2A$

Opening the bracket, we get

$\left\{ \tan \left( 3A \right)-\tan A\tan 2 \right.\left. A\tan \left( 3A \right) \right\}=\tan A+\tan 2A$

Now on rearranging the terms, by taking (tanA + tan2A) from right to left hand side and taking tanA tan2A tan3A from left to right hand side we get,

tan 3A – tan 2A – tan A = tan A tan 2A tan 3A.

Hence proved.

Note: Students while proving the identity get stuck after the $\dfrac{\sin (A+B)}{\cos (A+B)}$ part they forget how to divide cos A cos B to both numerator and denominator. Also, while doing cross multiplication be careful to avoid any calculation errors.

Another way to solve this is using the identity, $\tan \left( A \right)=\dfrac{\sin (A)}{\cos (A)}$. But this becomes a lengthy process.

Complete step-by-step answer:

For proving what is the asked in the question we have to use the identity

$\tan \left( A+B \right)=\dfrac{\tan A+tanB}{1-\tan A\tan B}$

Where A and B are any 2 angles.

First of all we have to prove the identity by using the facts.

\[\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B,cos\left( A+B \right)=\cos A\cos B-\sin A\sin B\]

Consider the identity,$\tan \left( A+B \right)=\dfrac{\sin (A+B)}{\cos (A+B)}$

Now we can write it as,

$\tan \left( A+B \right)=\dfrac{\text{sinA cosB+cosA sinB}}{\text{cosA cosB-sinA sinB}}$

Now dividing by $\cos A\cos B$ in numerator and denominator we get,

$\tan \left( A+B \right)=\dfrac{\dfrac{\text{sinA cosB}}{\text{cosA cosB}}\text{+}\dfrac{\text{cosA sinB}}{\text{cosA cosB}}}{\dfrac{\text{cosA cosB}}{\text{cosA cosB}}\text{ - }\dfrac{\text{sinA sinB}}{\text{cosA cosB}}}$

Cancelling the like terms, we get

$\tan \left( A+B \right)=\dfrac{\operatorname{tanA}+tanB}{\text{1-tanA tanB}}$

Now we can use the identity by taking ‘B’ as ‘2A’

Here by substituting B as 2A we get,

$\tan \left( A+2A \right)=\dfrac{\operatorname{tanA}+tan2A}{\text{1-tanA tan2A}}$

On further simplification we get,

$\tan \left( 3A \right)=\dfrac{\operatorname{tanA}+tan2A}{\text{1-tan2A tanA}}$

Now on cross multiplication we get,

$\tan \left( 3A \right)\left\{ 1-\tan A\tan 2 \right.\left. A \right\}=\tan A+\tan 2A$

Opening the bracket, we get

$\left\{ \tan \left( 3A \right)-\tan A\tan 2 \right.\left. A\tan \left( 3A \right) \right\}=\tan A+\tan 2A$

Now on rearranging the terms, by taking (tanA + tan2A) from right to left hand side and taking tanA tan2A tan3A from left to right hand side we get,

tan 3A – tan 2A – tan A = tan A tan 2A tan 3A.

Hence proved.

Note: Students while proving the identity get stuck after the $\dfrac{\sin (A+B)}{\cos (A+B)}$ part they forget how to divide cos A cos B to both numerator and denominator. Also, while doing cross multiplication be careful to avoid any calculation errors.

Another way to solve this is using the identity, $\tan \left( A \right)=\dfrac{\sin (A)}{\cos (A)}$. But this becomes a lengthy process.

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