Prove the trigonometric expression: \[\tan 3A\tan 2A\tan A=\tan 3Atan2AtanA\]
Answer
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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.
Last updated date: 17th Sep 2023
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