Answer
Verified
424.5k+ views
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.
Recently Updated Pages
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred
The branch of science which deals with nature and natural class 10 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Define absolute refractive index of a medium
Find out what do the algal bloom and redtides sign class 10 biology CBSE
Prove that the function fleft x right xn is continuous class 12 maths CBSE
Trending doubts
Change the following sentences into negative and interrogative class 10 english CBSE
Write an application to the principal requesting five class 10 english CBSE
What is the z value for a 90 95 and 99 percent confidence class 11 maths CBSE
Fill the blanks with proper collective nouns 1 A of class 10 english CBSE
Write a letter to the principal requesting him to grant class 10 english CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE
Select the word that is correctly spelled a Twelveth class 10 english CBSE
What is a collective noun for bees class 10 english CBSE