Question

(Sec^2A-4tan^2A)(3-4sin^2A) = (1-4sin^2A)(3-tan^2A). prove it

Answer 1

$$((\sec A)^2 - 4 (\tan A)^2) * (3 - 4 (\sin A)^2) = (1 - 4 (\sin A)^2) * (3 - (\tan A)^2)$$

Reveal the brackets:

$$\begin{array}{l} 3 (\sec A)^2 - 4 (\sin A)^2 * (\sec A)^2 - 12 (\tan A)^2 + 16 (\tan A)^2 * (\sin A)^2 \\ = 3 - (\tan A)^2 - 12 (\sin A)^2 + 4 (\sin A)^2 * (\tan A)^2 \end{array}$$

We know, that $(\sec A)^2 = \frac{1}{(\cos A)^2}$

So $(\sin A)^2 * (\sec A)^2 = \frac{(\sin A)^2}{(\cos A)^2} = (\tan A)^2$

$$\begin{array}{l} 3 (\sec A)^2 - 4 (\tan A)^2 - 12 (\tan A)^2 + 16 (\tan A)^2 * (\sin A)^2 \\ = 3 - (\tan A)^2 - 12 (\sin A)^2 + 4 (\sin A)^2 * (\tan A)^2 \end{array}$$ $$3 ((\sec A)^2 - 1) - 15 (\tan A)^2 = -12 (\sin A)^2 - 12 (\sin A)^2 * (\tan A)^2$$

As we know $1 + (\tan A)^2 = (\sec A)^2$

That is why $(\sec A)^2 - 1 = (\tan A)^2$

$$\begin{array}{l} 3 (\tan A)^2 - 15 (\tan A)^2 = -12 (\sin A)^2 * (1 + (\tan A)^2) \\ -12 (\tan A)^2 = -12 (\sin A)^2 * (\sec A)^2 \\ (\tan A)^2 = (\tan A)^2 \\ 1 = 1 \end{array}$$

Proved