Question

show that ( 1 - sin2 A ) ( 1 + tan 2 A ) = 1​

Answer 1

Answer:

​$( 1 - sin^2 A ) ( 1 + tan^2 A ) = 1$

Identities to know

$1-sin^2A = cos^2A$

$tan^2A = \frac{sin^2A}{cos^2A}$

$cos^2A +sin^2A = 1$

Rewrite equation in terms of sin and cos

$(1-sin^2A)(1+sin^2A/cos^2A)$

Now Multiply

$1 + (sin^2A/cos^2A) - sin^2A -(sin^4A/cos^2A)$

Rearrange  sin^2A to get the cos identity

$1-sin^2A +(sin^2A/cos^2A) - (sin^4A/cos^2A)$

$cos^2A +(sin^2A/cos^2A) - (sin^4A/cos^2A)$

combine terms with cos^2A as the denominator

$cos^2A +(sin^2A-sin^4A/cos^2A)$

Factor out sin

$cos^2A +sin^2A(1-sin^2A/cos^2A)$

$cos^2A +sin^2A(cos^2A/cos^2A)$

$cos^2A +sin^2A = 1$