Question

Sin^2 2a-4sin^2a/sin^2 2a+4sin^2a-4=tg^4 a

Answer 1

$\frac{Sin^{2}2\alpha-4Sin^{2}\alpha}{Sin^{2}2\alpha+4Sin^{2}\alpha-4} =\frac{(2Sin\alpha Cos\alpha)^{2}-4Sin^{2}\alpha}{(2Sin\alpha Cos\alpha)^{2}+(4Sin^{2}\alpha-4)} =\frac{4Sin^{2}\alpha Cos^{2} \alpha-4Sin^{2}\alpha }{4Sin^{2}\alpha Cos^{2}\alpha+4(Sin^{2}\alpha-1)}=\frac{4Sin^{2}\alpha(Cos^{2}\alpha-1)}{4Sin^{2}\alpha Cos^{2}\alpha-4Cos^{2}\alpha}=\frac{-4Sin^{2}\alpha Sin^{2}\alpha}{4Cos^{2}\alpha(Sin^{2}\alpha-1)}=\frac{-4Sin^{4}\alpha}{-4Cos^{4}\alpha}=tg^{4}\alpha$

Answer 2

To prove the trigonometric identity $\frac{\sin^2 2a - 4\sin^2 a}{\sin^2 2a + 4\sin^2 a - 4} = \tan^4 a$, we will start by simplifying the left-hand side of the equation step by step:

Given expression: $\frac{\sin^2 2a - 4\sin^2 a}{\sin^2 2a + 4\sin^2 a - 4}$

Step 1: Factor out $4$ from the numerator and denominator:

$\frac{4(\frac{\sin^2 2a}{4} - \sin^2 a)}{4(\frac{\sin^2 2a}{4} + \sin^2 a - 1)}$

Step 2: Notice that $\frac{\sin^2 2a}{4} = \frac{1 - \cos^2 2a}{4}$ (using the identity $\sin^2 x = 1 - \cos^2 x$):

$\frac{4(\frac{1 - \cos^2 2a}{4} - \sin^2 a)}{4(\frac{1 - \cos^2 2a}{4} + \sin^2 a - 1)}$

Step 3: Distribute $4$ in the numerator and denominator:

$\frac{1 - \cos^2 2a - 4\sin^2 a}{1 - \cos^2 2a + 4\sin^2 a - 4}$

Step 4: Use the double-angle identity for $\cos 2a$:

$\frac{1 - (2\cos^2 a - 1) - 4\sin^2 a}{1 - (2\cos^2 a - 1) + 4\sin^2 a - 4}$

Step 5: Simplify:

$\frac{2 - 2\cos^2 a - 4\sin^2 a}{-2\cos^2 a + 4\sin^2 a}$

Step 6: Apply the Pythagorean identity $\sin^2 a + \cos^2 a = 1$:

$\frac{2 - 2\cos^2 a - 4(1 - \cos^2 a)}{-2\cos^2 a + 4(1 - \cos^2 a)}$

Step 7: Simplify:

$\frac{2 - 2\cos^2 a - 4 + 4\cos^2 a}{-2\cos^2 a + 4 - 4\cos^2 a}$

Step 8: Combine like terms:

$\frac{2\cos^2 a - 2}{2 - 6\cos^2 a}$

Step 9: Divide both the numerator and denominator by $2$:

$\frac{\cos^2 a - 1}{1 - 3\cos^2 a}$

Step 10: Use the identity $\tan^2 a = \frac{\sin^2 a}{\cos^2 a}$