Question

4tan A/1-tan^4A=tan 2A+sin2A

Answer 1

Answer on Question #46771 – Mathematics – Trigonometry

**Question:**

Prove the trigonometric identity

$$\frac{4 \tan A}{1 - \tan^4 A} = \tan 2A + \sin 2A$$

**Solution:**

Let’s prove the given trigonometric identity in several steps.

1) Rewrite the denominator in a left side of identity as a product of two factors:

$$\frac{4 \tan A}{1 - \tan^4 A} = \frac{4 \tan A}{(1 - \tan^2 A)(1 + \tan^2 A)}$$

2) Use the Pythagorean identity $1 + \tan^2 A = \frac{1}{\cos^2 A}$ to simplify the expression in the denominator of (1):

$$\frac{4 \tan A}{(1 - \tan^2 A)(1 + \tan^2 A)} = \frac{4 \tan A \cdot \cos^2 A}{1 - \tan^2 A}$$

3) Use the tangent double-angle formula $\tan 2A = \frac{2 \tan A}{1 - \tan^2 A}$ to rewrite the right side of (2):

$$\frac{4 \tan A \cdot \cos^2 A}{1 - \tan^2 A} = \tan 2A \cdot 2 \cos^2 A.$$

4) Using the cosine power-reduction formula $\cos^2 A = \frac{1 + \cos^2 A}{2}$ and representing the tangent of an angle as the ratio of the sine to the cosine: $\tan A = \frac{\sin A}{\cos A}$, we obtain

$$\begin{aligned} \tan 2A \cdot 2 \cos^2 A &= \tan 2A \cdot 2 \left(\frac{1 + \cos^2 A}{2}\right) = \tan 2A \cdot (1 + \cos^2 A) = \tan 2A + \tan 2A \cdot \cos^2 A \\ &= \tan 2A + \frac{\sin^2 A}{\cos^2 A} \cdot \cos^2 A = \tan 2A + \sin^2 A. \end{aligned}$$

Thus, we prove that

$$\begin{aligned} \frac{4 \tan A}{1 - \tan^4 A} &= \frac{4 \tan A}{(1 - \tan^2 A)(1 + \tan^2 A)} = \frac{4 \tan A \cdot \cos^2 A}{1 - \tan^2 A} = \tan 2A \cdot 2 \cos^2 A = \tan 2A \cdot 2 \left(\frac{1 + \cos^2 A}{2}\right) \\ &= \tan 2A \cdot (1 + \cos^2 A) = \tan 2A + \tan 2A \cdot \cos^2 A = \tan 2A + \frac{\sin^2 A}{\cos^2 A} \cdot \cos^2 A \\ &= \tan 2A + \sin^2 A, \end{aligned}$$

i.e.

$$\frac{4 \tan A}{1 - \tan^4 A} = \tan 2A + \sin 2A$$

**Answer:** The given trigonometric identity is proved.

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