Question

1+tanx 1–tanx 2 = tan 4 tan 4 x x π π   +       −  .

Answer 1

The given trigonometric identity is proved to be true.

Answer 2

We start with the well‐known formula

$$\frac{1+\tan x}{1-\tan x}=\tan\Bigl(x+\frac{\pi}{4}\Bigr).$$

Squaring both sides gives

$$\left(\frac{1+\tan x}{1-\tan x}\right)^2=\tan^2\Bigl(x+\frac{\pi}{4}\Bigr).$$

Now note that

$$\tan\Bigl(\frac{\pi+4x}{4}\Bigr)=\tan\Bigl(x+\frac{\pi}{4}\Bigr)$$

and

$$\tan\Bigl(\frac{\pi-4x}{4}\Bigr)=\tan\Bigl(\frac{\pi}{4}-x\Bigr).$$

Thus, writing the right‐side of the given identity as

$$\frac{\tan\Bigl(\frac{\pi+4x}{4}\Bigr)}{\tan\Bigl(\frac{\pi-4x}{4}\Bigr)} =\frac{\tan\Bigl(x+\frac{\pi}{4}\Bigr)}{\tan\Bigl(\frac{\pi}{4}-x\Bigr)},$$

we use the property that the angles $x+\frac{\pi}{4}$ and $\frac{\pi}{4}-x$ are complementary (since

$$\Bigl(x+\frac{\pi}{4}\Bigr)+\Bigl(\frac{\pi}{4}-x\Bigr)=\frac{\pi}{2},$$

and recall that if $\theta$ and $\frac{\pi}{2}-\theta$ are complementary then

$$\tan \theta \cdot \tan\Bigl(\frac{\pi}{2}-\theta\Bigr)=1,$$

or equivalently,

$$\tan\Bigl(\frac{\pi}{2}-\theta\Bigr)=\frac{1}{\tan \theta}\,).$$

Taking $\theta=x+\frac{\pi}{4}$ we obtain:

$$\tan\Bigl(\frac{\pi}{4}-x\Bigr)=\cot\Bigl(x+\frac{\pi}{4}\Bigr)=\frac{1}{\tan\Bigl(x+\frac{\pi}{4}\Bigr)}.$$

Thus,

$$\frac{\tan\Bigl(x+\frac{\pi}{4}\Bigr)}{\tan\Bigl(\frac{\pi}{4}-x\Bigr)} =\tan\Bigl(x+\frac{\pi}{4}\Bigr)\cdot \tan\Bigl(x+\frac{\pi}{4}\Bigr) =\tan^2\Bigl(x+\frac{\pi}{4}\Bigr).$$

Since both the left‐side and the transformed right‐side are equal to $\tan^2\Bigl(x+\frac{\pi}{4}\Bigr)$, the identity

$$\left(\frac{1+\tan x}{1-\tan x}\right)^2=\frac{\tan\Bigl(\frac{\pi+4x}{4}\Bigr)}{\tan\Bigl(\frac{\pi-4x}{4}\Bigr)}$$

is verified.