Question

Sin2 A. Cos2B+cos2 A. Sin2B+ cos2A. Cos2 B+sin2A. Sin2B=1​

Answer 1

Answer:

$\sin^2 A.cos^2B+cos^2A.sin^2B+ cos^2A. cos^2B+sin^2A.sin^2B=1$

Step-by-step explanation:

$\text{Prove that:}sin^2 A.cos^2B+cos^2A.sin^2B+ cos^2A. cos^2B+sin^2A.sin^2B=1$

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

using

$\boxed{cos^2\theta=1-sin^2\theta}$

$=sin^2 A.(1-sin^2B)+(1-sin^2A).sin^2B+(1-sin^2A) (1-sin^2B)+sin^2A.sin^2B$

$=sin^2 A-sin^2A\:sin^2B+sin^2B-sin^2A\:sin^2B+(1-sin^2A) (1-sin^2B)+sin^2A\:sin^2B$

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$=sin^2 A+sin^2B-sin^2A\:sin^2B+(1-sin^2A) (1-sin^2B)$

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$=sin^2 A+sin^2B-sin^2A\:sin^2B+1-sin^2A-sin^2B+sin^2A\:sin^2B$

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$=sin^2 A+sin^2B+1-sin^2A-sin^2B$

​

$=1$

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$\implies\:\boxed{sin^2 A.cos^2B+cos^2A.sin^2B+ cos^2A. cos^2B+sin^2A.sin^2B=1}$

Answer 2

Answer:

hope it helps!!!!!

Step-by-step explanation:

sin2acos2b+cos2asin2b+cos2acos2b+sin2asin2b=1