If tan|+sin|=m and tan|- sin|=n, show that (m2-n2)=4?mn

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m =tanA + sinA n =tanA - sinA m^{2}-n^{2} =(tanA + sinA)^{2} - (tanA-sinA)^{2} m^{2}-n^{2} =4tanA.sinA mn = (tanA+sinA)(tanA-sinA) mn = tan^{2}A-sin^{2}A mn = sin^{2}A(\frac{1}{cos^{2}A}-1) mn = sin^{2}A(\frac{1-cos^{2}A}{cos^{2}A}) mn = sin^{2}A(\frac{sin^{2}A}{cos^{2}A}) mn = sin^{2}A.tan^{2}A \sqrt{mn}...
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m =tanA + sinA n =tanA - sinA m^{2}-n^{2} =(tanA + sinA)^{2} - (tanA-sinA)^{2} m^{2}-n^{2} =4tanA.sinA mn = (tanA+sinA)(tanA-sinA) mn = tan^{2}A-sin^{2}A mn = sin^{2}A(\frac{1}{cos^{2}A}-1) mn = sin^{2}A(\frac{1-cos^{2}A}{cos^{2}A}) mn = sin^{2}A(\frac{sin^{2}A}{cos^{2}A}) mn = sin^{2}A.tan^{2}A \sqrt{mn} = sinA.tanA m^{2}-n^{2} =4tanA.sinA m^{2}-n^{2} =4\sqrt{mn} read less
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