Question

If tan^2 alpha= cos^2 beta-sin^2beta, then prove cos^2alpha-sin^2alpha= tan^2 beta

Answer 1

Answer:

proved

Step-by-step explanation:

Given If tan^2 alpha= cos^2 beta-sin^2beta, then prove cos^2alpha-sin^2alpha= tan^2 beta

We need to know the formula

cos2x = 1 - tan^2 x /1 + tan ^2 x

cos2x = cos^2 x - sin ^2 x

Now it is given

tan ^2 alpha = cos^2 beta - sin^2 beta

tan^2 alpha = cos2 beta

tan^2 alha = 1 - tan^2 beta / 1 + tan^2 beta

By componendo and dividendo we get

1 - tan^2 alpha/1 + tan^2 alpha = 1 + tan^2 beta - (1 - tan^2 beta) / 1 + tan^2        beta + 1 - tan^2 beta

cos2 alpha = 2 tan^2 beta / 2

tan^2 beta = cos^2 alpha - sin^2 alpha

Answer 2

Answer:

$cos^2\alpha-sin^2\alpha=tan^2\beta$

Step-by-step explanation:

Formula used:

$1.cos2A=cos^2A-sin^2A\\\\2.cos2A=\frac{1-tan^2A}{1+tan^2A}$

Given:

$tan^2\alpha=cos^2\beta-sin^2\beta$

Now,

By formula(2)

$cos2\alpha=\frac{1-tan^2\alpha}{1+tan^2\alpha}\\\\cos2\alpha=\frac{(cos^2\beta+sin^2\beta)-(cos^2\beta-sin^2\beta)}{(cos^2\beta+sin^2\beta)+(cos^2\beta-sin^2\beta)}\\\\cos2\alpha=\frac{cos^2\beta+sin^2\beta-cos^2\beta+sin^2\beta}{cos^2\beta+sin^2\beta+cos^2\beta-sin^2\beta}\\\\cos2\alpha=\frac{2sin^2\beta}{2cos^2\beta}\\\\cos2\alpha=\frac{sin^2\beta}{cos^2\beta}\\\\cos2\alpha=tan^2\beta$

By formula(1)

$cos^2\alpha-sin^2\alpha=tan^2\beta$