Tan squared theta minus sin squared theta barabar 10 squared theta into sin squared theta prove that
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This formula is the Pythagorean theorem in disguise.
{\displaystyle \sin ^{2}(\theta )+\cos ^{2}(\theta )=1}
If you look at the diagram in the next section it should be clear why. Sine and Cosine of an angle {\displaystyle \theta } in a triangle with unit hypotenuse are just the lengths of the two shorter sides. So squaring them and adding gives the hypotenuse squared, which is one squared, which is one.
{\displaystyle \sin ^{2}(\theta )+\cos ^{2}(\theta )=1}
If you look at the diagram in the next section it should be clear why. Sine and Cosine of an angle {\displaystyle \theta } in a triangle with unit hypotenuse are just the lengths of the two shorter sides. So squaring them and adding gives the hypotenuse squared, which is one squared, which is one.
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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.
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