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Trignometry

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4 Apr 2008 22:35:03 IST

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Trigonometry question.

None

Show that 1+sin A cos B

----------------- + ------------------- =2(sin A- sin B)

cos A 1-sin B -----------------------------

sin(A-B)+(cos A- cos B)

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sreeraman nagasubramaniyan

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4 Apr 2008 23:47:53 IST

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$\mbox{LHS:} \ \frac{1+\sin(A)}{\cos(A)}+\frac{\cos(B)}{1-\sin(B)} \\ \\ = \frac{1+\sin(A)}{\cos(A)}+\frac{1+\sin(B)}{\cos(B)} \\ \\ =\sec(A)+\sec(B)+\tan(A)+\tan(B) \\ \\ = \frac{\cos(A)+\cos(B)+\sin(A+B)}{\cos(A)\cos(B)} \\ \\ \\ \frac{2\cos(\frac{A+B}{2})[\cos(\frac{A-B}{2})+\sin(\frac{A+B}{2})]}{\cos(A)\cos(B)} \\ \\ \\ \mbox{RHS:} \ \frac{2(\sin(A)-\sin(B))}{\sin(A-B)+\cos(A)-\cos(B)} \\ \\ \\ \frac{4(\cos(\frac{A+B}{2})\sin(\frac{A-B}{2}))}{2\sin(\frac{A-B}{2})[\cos(\frac{A-B}{2})-\sin(\frac{A+B}{2})]} \\ \\ \\ =\frac{2\cos(\frac{A+B}{2})}{\cos(\frac{A-B}{2})-\sin(\frac{A+B}{2})} \\ \\ \\ = \frac{2\cos(\frac{A+B}{2})}{\cos(\frac{A-B}{2})-\sin(\frac{A+B}{2})}* \frac{\cos(\frac{A-B}{2})+\sin(\frac{A+B}{2})}{\cos(\frac{A-B}{2})+\sin(\frac{A+B}{2})} \\ \\ \\(\cos(\frac{A-B}{2})-\sin(\frac{A+B}{2}))*(\cos(\frac{A-B}{2})+\sin(\frac{A+B}{2})) \\ \\= \cos^2(\frac{A-B}{2})-\sin^2(\frac{A+B}{2}) \\ \\ \cos^2(A)-\sin^2(B) =\cos(A+B)\cos(A-B) \\ \\ \mbox{Thus} \ \cos^2(\frac{A-B}{2})-\sin^2(\frac{A+B}{2}) = \cos(A)\cos(B) \\ \\ \mbox{Thus RHS is} \ \frac{2\cos(\frac{A+B}{2})[\cos(\frac{A-B}{2})+\sin(\frac{A+B}{2})]}{\cos(A)\cos(B)} \\ \\ \mbox{LHS = RHS,hence proved}$

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