If tan(π/4 + x) + tan(π/4 − x) = a, Find tan²(π/4 + x) + tan²(π/4 − x)

Solution

Let

\[ p=\tan\left(\frac{\pi}{4}+x\right) \]

and

\[ q=\tan\left(\frac{\pi}{4}-x\right) \]

Given,

\[ p+q=a \]

Now,

\[ pq = \tan\left(\frac{\pi}{4}+x\right) \tan\left(\frac{\pi}{4}-x\right) \]

Using the identity:

\[ \tan\left(\frac{\pi}{4}+x\right) = \frac{1+\tan x}{1-\tan x} \]

and

\[ \tan\left(\frac{\pi}{4}-x\right) = \frac{1-\tan x}{1+\tan x} \]

Therefore,

\[ pq=1 \]

Now use:

\[ p^2+q^2=(p+q)^2-2pq \]

Substituting values,

\[ p^2+q^2=a^2-2(1) \]

\[ =a^2-2 \]

Therefore,

\[ \boxed{ \tan^2\left(\frac{\pi}{4}+x\right) + \tan^2\left(\frac{\pi}{4}-x\right) = a^2-2 } \]

Final Answer

\[ \boxed{ \tan^2\left(\frac{\pi}{4}+x\right) + \tan^2\left(\frac{\pi}{4}-x\right) = a^2-2 } \]

Correct Option: (c)