Question

Evaluate $\tan \left( {\dfrac{\pi }{4} + \theta } \right) - \tan \left( {\dfrac{\pi }{4} - \theta } \right) =$
A. $2\tan 2\theta$
B. $2\cot \theta$
C. $\tan 2\theta$
D. $\cot 2\theta$

Answer 1

We can solve this using the tangent sum and difference formula. That is we have tangent sum formula $\tan \left( {A + B} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A.\tan B}}$ and tangent difference formula $\tan \left( {A - B} \right) = \dfrac{{\tan A - \tan B}}{{1 + \tan A.\tan B}}$. After applying the formula we use algebraic identities ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$ and ${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$ to simplify it.

Complete step by step answer:
Given $\tan \left( {\dfrac{\pi }{4} + \theta } \right) - \tan \left( {\dfrac{\pi }{4} - \theta } \right) - - - (1)$
Now take $\tan \left( {\dfrac{\pi }{4} + \theta } \right)$ and applying tangent sum formula that is $\tan \left( {A + B} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A.\tan B}}$.
Then
$\Rightarrow \tan \left( {\dfrac{\pi }{4} + \theta } \right) = \dfrac{{\tan \left( {\dfrac{\pi }{4}} \right) + \tan \theta }}{{1 - \tan \left( {\dfrac{\pi }{4}} \right).\tan \theta }}$
We know $\tan \left( {\dfrac{\pi }{4}} \right) = 1$ then
$\Rightarrow \tan \left( {\dfrac{\pi }{4} + \theta } \right) = \dfrac{{1 + \tan \theta }}{{1 - \tan \theta }} - - - (2)$

Now similarly take $\tan \left( {\dfrac{\pi }{4} - \theta } \right)$ and apply the tangent difference formula that is $\tan \left( {A - B} \right) = \dfrac{{\tan A - \tan B}}{{1 + \tan A.\tan B}}$. Then we have
$\Rightarrow \tan \left( {\dfrac{\pi }{4} - \theta } \right) = \dfrac{{\tan \left( {\dfrac{\pi }{4}} \right) - \tan \theta }}{{1 + \tan \left( {\dfrac{\pi }{4}} \right).\tan \theta }}$
We know $\tan \left( {\dfrac{\pi }{4}} \right) = 1$ then
$\Rightarrow \tan \left( {\dfrac{\pi }{4} - \theta } \right) = \dfrac{{1 - \tan \theta }}{{1 + \tan \theta }} - - - - (3)$
Now substituting equation (2) and (3) in (1) we have,
$\tan \left( {\dfrac{\pi }{4} + \theta } \right) - \tan \left( {\dfrac{\pi }{4} - \theta } \right) = \dfrac{{1 + \tan \theta }}{{1 - \tan \theta }} - \dfrac{{1 - \tan \theta }}{{1 + \tan \theta }}$
$\Rightarrow \tan \left( {\dfrac{\pi }{4} + \theta } \right) - \tan \left( {\dfrac{\pi }{4} - \theta } \right)= \dfrac{{1 + \tan \theta }}{{1 - \tan \theta }} - \dfrac{{1 - \tan \theta }}{{1 + \tan \theta }}$

Taking LCM and simplifying we have,
$\tan \left( {\dfrac{\pi }{4} + \theta } \right) - \tan \left( {\dfrac{\pi }{4} - \theta } \right) = \dfrac{{\left[ {\left( {1 + \tan \theta } \right)\left( {1 + \tan \theta } \right)} \right] - \left[ {\left( {1 - \tan \theta } \right)\left( {1 - \tan \theta } \right)} \right]}}{{\left( {1 - \tan \theta } \right)\left( {1 + \tan \theta } \right)}}$
We know the $(a - b)(a + b) = {a^2} - {b^2}$.
$\tan \left( {\dfrac{\pi }{4} + \theta } \right) - \tan \left( {\dfrac{\pi }{4} - \theta } \right)= \dfrac{{\left[ {{{\left( {1 + \tan \theta } \right)}^2}} \right] - \left[ {{{\left( {1 - \tan \theta } \right)}^2}} \right]}}{{{1^2} - {{\tan }^2}\theta }}$
Now we use ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$ and ${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$, then
$\Rightarrow \tan \left( {\dfrac{\pi }{4} + \theta } \right) - \tan \left( {\dfrac{\pi }{4} - \theta } \right) = \dfrac{{\left[ {{1^2} + {{\tan }^2}\theta + 2\tan \theta } \right] - \left[ {{1^2} + {{\tan }^2}\theta - 2\tan \theta } \right]}}{{{1^2} - {{\tan }^2}\theta }}$

Expanding the brackets we have,
$\tan \left( {\dfrac{\pi }{4} + \theta } \right) - \tan \left( {\dfrac{\pi }{4} - \theta } \right)= \dfrac{{1 + {{\tan }^2}\theta + 2\tan \theta - 1 - {{\tan }^2}\theta + 2\tan \theta }}{{1 - {{\tan }^2}\theta }}$
$\Rightarrow \tan \left( {\dfrac{\pi }{4} + \theta } \right) - \tan \left( {\dfrac{\pi }{4} - \theta } \right) = \dfrac{{2\tan \theta + 2\tan \theta }}{{1 - {{\tan }^2}\theta }}$
$\Rightarrow \tan \left( {\dfrac{\pi }{4} + \theta } \right) - \tan \left( {\dfrac{\pi }{4} - \theta } \right) = \dfrac{{4\tan \theta }}{{1 - {{\tan }^2}\theta }}$.
This is the simplified form but it is not matching with the given option. So we need to simply this further,
$\tan \left( {\dfrac{\pi }{4} + \theta } \right) - \tan \left( {\dfrac{\pi }{4} - \theta } \right) = \dfrac{{2 \times \left( {2\tan \theta } \right)}}{{1 - {{\tan }^2}\theta }}$

We can write $2\tan \theta = \tan \theta + \tan \theta$ and also ${\tan ^2}\theta = \tan \theta \tan \theta$.
$\Rightarrow \tan \left( {\dfrac{\pi }{4} + \theta } \right) - \tan \left( {\dfrac{\pi }{4} - \theta } \right) = \dfrac{{2 \times \left( {\tan \theta + \tan \theta } \right)}}{{1 - \tan \theta .\tan \theta }}$
If we observe we have a tangent sum rule,
$\tan \left( {\dfrac{\pi }{4} + \theta } \right) - \tan \left( {\dfrac{\pi }{4} - \theta } \right) = 2 \times \left( {\dfrac{{\left( {\tan \theta + \tan \theta } \right)}}{{1 - \tan \theta .\tan \theta }}} \right)$
$\Rightarrow \tan \left( {\dfrac{\pi }{4} + \theta } \right) - \tan \left( {\dfrac{\pi }{4} - \theta } \right) = 2 \times \tan \left( {\theta + \theta } \right)$
$\therefore \tan \left( {\dfrac{\pi }{4} + \theta } \right) - \tan \left( {\dfrac{\pi }{4} - \theta } \right) = 2\tan \left( {2\theta } \right)$.

Hence the correct answer is option A.

Note: We also have sine and cosine sum and difference formulas. We have sine sum and difference formula that is $\sin \left( {A + B} \right) = \sin A.\cos B + \cos A.\sin B$ and $\sin \left( {A - B} \right) = \sin A.\cos B - \cos A.\sin B$. The cosine sum and difference formula is $\cos (A + B) = \cos A.\cos B - \sin A.\sin B$ and $\cos (A - B) = \cos A.\cos B + \sin A.\sin B$. We use them depending on the given problem. In above while expanding the brackets we have to be careful regarding the negative sign. We know that the product of two negative numbers results in a positive number. Also the product of positive (negative) numbers and negative (positive) numbers results in negative numbers.