Question

How can you prove the trigonometric identity of the following: $\tan (x + {45^ \circ }) - \tan ({45^ \circ } - x) = 2\tan 2x$

Answer 1

Hint: We can solve these types of questions easily if we are familiar with its formulas. We will expand the two terms of tan which are on the left side. We will take the L.C.M. of the denominator and simplify them. We will use formulas and try to convert the equation in terms of tan2x.

Complete step-by-step solution:
We have to prove that $\tan (x + {45^ \circ }) - \tan ({45^ \circ } - x)$ is equal to $2\tan 2x$ .
We will firstly simplify the term $\tan (x + {45^ \circ })$ and $\tan ({45^ \circ } - x)$ individually
We know that \[\;tan(a + b) = \dfrac{{tana + tanb}}{{1 - tana \times tanb}}\] ……………(1)
 We will use the equation 1 and expand $\tan (x + {45^ \circ })$
\[\; \Rightarrow tan(x + 45) = \dfrac{{\tan x + tan45}}{{1 - \tan x \times tan45}}\]
We have put the value of tan45
\[\; \Rightarrow tan(x + 45) = \dfrac{{\tan x + 1}}{{1 - \tan x \times 1}}\] ………………(2)
We know that \[\;tan(a - b) = \dfrac{{tana - tanb}}{{1 + tana \times tanb}}\] ……………(3)
Similarly, we will expand $\tan ({45^ \circ } - x)$ using the equation 3
\[\; \Rightarrow tan(45 - x) = \dfrac{{\tan 45 - \tan x}}{{1 + \tan 45 \times \tan x}}\]
We have put the value of tan45
\[\; \Rightarrow tan(45 - x) = \dfrac{{1 - \tan x}}{{1 + 1 \times \tan x}}\] ………………….(4)
We will substitute the value from the equation 2 and 4 in the equation $\tan (x + {45^ \circ }) - \tan ({45^ \circ } - x)$
\[ = \tan (x + {45^ \circ }) - \tan ({45^ \circ } - x)\]
\[ = \;\dfrac{{\tan x + 1}}{{1 - \tan x}} - \dfrac{{1 - \tan x}}{{1 + \tan x}}\]
We will take the L.C.M. and simplify it
\[ = \;\dfrac{{{{\left( {\tan x + 1} \right)}^2} - {{\left( {1 - \tan x} \right)}^2}}}{{\left( {1 - \tan x} \right)\left( {1 + \tan x} \right)}}\]
We have expanded the squared terms
\[ = \;\dfrac{{{{\tan }^2}x + 1 + 2\tan x - \left( {1 + {{\tan }^2}x - 2\tan x} \right)}}{{\left( {1 - \tan x} \right)\left( {1 + \tan x} \right)}}\]
\[ = \;\dfrac{{{{\tan }^2}x + 1 + 2\tan x - 1 - {{\tan }^2}x + 2\tan x}}{{\left( {1 - \tan x} \right)\left( {1 + \tan x} \right)}}\]
\[ = \;\dfrac{{4\tan x}}{{\left( {1 - \tan x} \right)\left( {1 + \tan x} \right)}}\]
We know that $\tan 2x = \dfrac{{2\tan x}}{{1 - {{\tan }^2}x}}$
So, we convert the above equation in terms of tan2x, we get
\[ = \;2\tan 2x\]
Hence, we proved that $\tan (x + {45^ \circ }) - \tan ({45^ \circ } - x) = 2\tan 2x$ .

Note: Sometimes we solve the question but we can't reach our final answer. It happens maybe because of the wrong approach, for this particular question another possibility is that we can't convert our last result to our desired result. We should also memorise all of the usual angle values of sin, cosine, and tan.