Question

If we have an expression $\tan \left( \dfrac{\pi }{4}+x \right)+\tan \left( \dfrac{\pi }{4}-x \right)=a$ , then ${{\tan }^{2}}\left( \dfrac{\pi }{4}+x \right)+{{\tan }^{2}}\left( \dfrac{\pi }{4}-x \right)=$
A. ${{a}^{2}}+1$
B. ${{a}^{2}}+2$
C. ${{a}^{2}}-2$
D. None of these

Answer 1

Hint: Here we have been given an equation of trigonometric function whose value is given using which we have to find the value of the term given. Firstly as we have to find the square value we will square both sides of the equation given. Then by using the sum identity of the tangent function we will find the value of some terms. Finally we will simplify the value obtained and get the desired answer.

Complete step-by-step solution:
It has been given to us that,
$\tan \left( \dfrac{\pi }{4}+x \right)+\tan \left( \dfrac{\pi }{4}-x \right)=a$……$\left( 1 \right)$
We have to find the value of,
${{\tan }^{2}}\left( \dfrac{\pi }{4}+x \right)+{{\tan }^{2}}\left( \dfrac{\pi }{4}-x \right)$
Firstly squaring both sides of the equation (1) we get,
${{\left( \tan \left( \dfrac{\pi }{4}+x \right)+\tan \left( \dfrac{\pi }{4}-x \right) \right)}^{2}}={{a}^{2}}$
Using the formula ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ above we get,
${{\tan }^{2}}\left( \dfrac{\pi }{4}+x \right)+{{\tan }^{2}}\left( \dfrac{\pi }{4}-x \right)+2\tan \left( \dfrac{\pi }{4}+x \right)\times \tan \left( \dfrac{\pi }{4}-x \right)={{a}^{2}}$…..$\left( 2 \right)$
Now as we know the sum identity of tangent is as follows:
\[\tan \left( a+b \right)=\dfrac{\tan a+\tan b}{1-\tan a\tan b}\] , \[\tan \left( a-b \right)=\dfrac{\tan a-\tan b}{1+\tan a\tan b}\]
Using it in equation (2) we get,
${{\tan }^{2}}\left( \dfrac{\pi }{4}+x \right)+{{\tan }^{2}}\left( \dfrac{\pi }{4}-x \right)+2\dfrac{\tan \dfrac{\pi }{4}+\tan x}{1-\tan \dfrac{\pi }{4}\tan x}\times \dfrac{\tan \dfrac{\pi }{4}-\tan x}{1+\tan \dfrac{\pi }{4}\tan x}={{a}^{2}}$
We know that $\tan \dfrac{\pi }{4}=1$ using it above we get,
$\Rightarrow {{\tan }^{2}}\left( \dfrac{\pi }{4}+x \right)+{{\tan }^{2}}\left( \dfrac{\pi }{4}-x \right)+2\dfrac{1+\tan x}{1-\tan x}\times \dfrac{1-\tan x}{1+\tan x}={{a}^{2}}$
$\Rightarrow {{\tan }^{2}}\left( \dfrac{\pi }{4}+x \right)+{{\tan }^{2}}\left( \dfrac{\pi }{4}-x \right)+2={{a}^{2}}$
Taking $2$ on right side we get,
${{\tan }^{2}}\left( \dfrac{\pi }{4}+x \right)+{{\tan }^{2}}\left( \dfrac{\pi }{4}-x \right)={{a}^{2}}-2$
So we got the value of ${{\tan }^{2}}\left( \dfrac{\pi }{4}+x \right)+{{\tan }^{2}}\left( \dfrac{\pi }{4}-x \right)$ as ${{a}^{2}}-2$ .
Hence the correct option is (C).

Note: In this type of question we should firstly see the highest power of the value we have to find then accordingly square or cube the value already given to us. Second step will be to remove the term whose value is not asked or whose value we can easily find. The calculation part is very important in such questions. Also one should know all the basic as well as important identities of the trigonometric functions. Trigonometry is a very important branch of mathematics. We have six basic trigonometric functions such as sine, cosine, tangent, cosecant, secant and cotangent.