proving

**lezah** · April 20th 2010, 08:37 PM

can you prove that:

1/(cos^2 x-sin^2 x)^2 - 4tan^2 x/ (1-tan^2 x)^2 =1

im kinda stuck right now.

**harish21** · April 20th 2010, 09:14 PM

Originally Posted by lezah

can you prove that:

1/(cos^2 x-sin^2 x)^2 - 4tan^2 x/ (1-tan^2 x)^2 =1

im kinda stuck right now.

$\frac{1}{(cos^2 x-sin^2 x)^2} - \frac{4tan^2 x}{(1-tan^2 x)^2}$

$= \frac{1}{(cos^2 x-sin^2 x)^2} - \frac{\frac{4sin^2x}{cos^2x}}{(1-\frac{sin^2x}{cos^2x})^2}$

$= \frac{1}{(cos^2 x-sin^2 x)^2} - \frac{\frac{4sin^2x}{cos^2x}}{\frac{(cos^2x-sin^2x)^2}{(cos^2x)^2}}$

$= \frac{1}{(cos^2 x-sin^2 x)^2} - (\frac{4sin^2x}{cos^2x} \times \frac{cos^4x}{(cos^2 x-sin^2 x)^2})$

$= \frac{1}{(cos^2 x-sin^2 x)^2} - \frac{4sin^2x cos^2x}{(cos^2 x-sin^2 x)^2}$

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

now expand $(cos^2 x-sin^2 x)^2$ as $cos^4x+sin^4x -2cos^2x. sin^2x$ $= (sin^2x)^2 + (cos^2x)^2-2cos^2x. sin^2x=(sin^2x+cos^2x)^2-2cos^2x. sin^2x-2cos^2x. sin^2x$

**sa-ri-ga-ma** · April 20th 2010, 11:13 PM

Originally Posted by lezah

can you prove that:

1/(cos^2 x-sin^2 x)^2 - 4tan^2 x/ (1-tan^2 x)^2 =1

im kinda stuck right now.

cos^2(x) - sin^2(x) = cos(2x)
2tax(x)/(1-tan^(x)] = tan(2x)
So the given problem is
1/cos^2(2x) - tan^2(2x)
sec^2(2x) - tan^2(2x) = .......?

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