Question

(1-sin^4(a)-cos^4(a))*(1+tg^2(a))=2sin^2(a)

Answer 1

Упростим левую часть:

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

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

$=(1+sin^2a)-cos^2a=(1-cos^2a)+sin^2a=sin^2a+sin^2a=$

$=2sin^2a$

Таким образом, левая часть равна правой. Доказано.

Answer 2

1+tg^2(a)=1/cos(a)^2

2sin^2(a)=1-cos(2*a)

(1-sin^4(a)-cos^4(a))=2*cos(a)^2-2*cos(a)^4

(2*cos(a)^2-2*cos(a)^4)*(1/cos(a)^2)=1-cos(2*a)

(2*cos(a)^2-2*cos(a)^4)*(1/cos(a)^2)-1-cos(2*a)=0

(1+sin^2(a)-cos^2(a)=(1-cos^2(a)+sin^2(a)

sin^2(a)+sin^2(a)=2sin^2(a)

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