Question

prove the following identities tan0/1-cot0 + cot0/1-tan0 = 1+sec0 cosec0

Answer 1

LHS = tanO/(1-cotO) + cotO/(1-tanO)
= $\frac{ \frac{sinO}{cosO} }{1- \frac{cosO}{sinO} } +\frac{ \frac{cosO}{sinO} }{1- \frac{sinO}{cosO} } \\ =\frac{ \frac{sinO}{cosO} }{ \frac{sinO-cosO}{sinO} } +\frac{ \frac{cosO}{sinO} }{ \frac{cosO-sinO}{cosO} } \\ = \frac{sin^2O}{cosO(sinO-cosO)} + \frac{cos^2O}{sinO(cosO-sinO)} \\ = \frac{sin^2O}{cosO(sinO-cosO)} - \frac{cos^2O}{sinO(sinO-cosO)} \\ = \frac{sin^3O-cos^3O}{sinOcosO(sinO-cosO)} \\ =\frac{(sinO-cosO)(sin^2O+cos^2O+sinOcosO)}{sinOcosO(sinO-cosO)} \\ = \frac{1+sinOcosO}{sinOcosO}$
= [1 +(1/cosecOsecO)]/[1/cosecOsecO]
=[(cosecOsecO+1)/cosecOsecO]*[cosecOsecO]
=cosecOsecO+1
=RHS

Hence proved

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Answer 2

Answer:

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