Question

prove The Identity sin square + cos square = 1​

Answer 1

it is the solution of your answer

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Answer 2

Consider the provided information.

\sin^2A\cos^2B-\cos^2A\sin^2B=\sin^2A-\sin^2B

Consider the LHS.

\sin^2A\cos^2B-\cos^2A\sin^2B

\sin^2A(1-\sin^2B)-(1-\sin^2A)\sin^2B               (∴\cos^2x=1-\sin^2x)

\sin^2A-\sin^2A\sin^2B-\sin^2B+\sin^2A\sin^2B

\sin^2A-\sin^2B

Hence, proved.

Consider the provided information.

\sin^2A\cos^2B-\cos^2A\sin^2B=\sin^2A-\sin^2B

Consider the LHS.

\sin^2A\cos^2B-\cos^2A\sin^2B

\sin^2A(1-\sin^2B)-(1-\sin^2A)\sin^2B               (∴\cos^2x=1-\sin^2x)

\sin^2A-\sin^2A\sin^2B-\sin^2B+\sin^2A\sin^2B

\sin^2A-\sin^2B

Hence, proved.