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Find out mathematical flaw...

Que: Prove that sin^4 A - sin^2 A cos^2 A +cos^4 A + 3 sin^2 A cos^2 A - (sin^2 A + cos^2 A) + sin^3 A cosec^2 A = sin A. Student: Multiply R.H.S. & L.H.S. by 0 i.e. 0=0. Teacher gave marks: 10/10 * 0 = 0. For you: Can you find out the mathematical flaw in the answer given by the student and the marks given by the teacher which have given us misleading results? NOTE: We often multiply both sides by a rational no. in such questions but if we do that with 0 it becomes misleading. What's the reason?

Note by Sarthak Kaushik
1 year, 10 months ago

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But 0 is not -ve. Then how can you apply the concept of squaring to this problem. Sarthak Kaushik · 1 year, 9 months ago

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\[\sin^4 A - \sin^2 A \cos^2 A +\cos^4 A + 3 \sin^2 A \cos^2 A - (\sin^2 A + \cos^2 A) + \sin^3 A \csc^2 A = \sin A\] \[(\sin^2A)^2+(\cos^2A)^2+ 2 \sin^2 A \cos^2 A - 1 + \sin^3 A \csc^2 A=\sin A\] \[(\sin^2A+cos^2A)^2-1+\sin^3 A \csc^2 A=\sin A\] \[\sin^3A\csc^2A=\sin A\] \[\sin^3A*\frac{1}{\sin^2A}=\sin A\] \[\sin A=\sin A\] Marc Vince Casimiro · 1 year, 10 months ago

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I think the reason why multiplying by \(0\) is misleading is that you cannot go back, i.e. dividing by \(0\) is not allowed. That follows the same concept when squaring a number. For example, given a negative number, say \(-a\) for some positive real number \(a^2\). If you square it, it would become \(a\) but you cannot say that getting the square root of \(a^2\) will lead you to \(a=-a\). I will post later for the proof and for those who cannot understand, it says: \[\sin^4 A - \sin^2 A \cos^2 A +\cos^4 A + 3 \sin^2 A \cos^2 A - (\sin^2 A + \cos^2 A) + \sin^3 A \csc^2 A = \sin A\] Hint: Sum of Two Squares and Pythagorean Theorem \[\] P.S. I thought that student was smart but then I realized that the teacher was smarter. :P Marc Vince Casimiro · 1 year, 10 months ago

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