Question

prove that : (image)​

Answer 1

LHS = tanO/(1-cotO) + cotO/(1-tanO= $\begin{gathered} \frac{ \frac{sinO}{cosO} }{1- \frac{cosO}{sinO} } +\frac{ \frac{cosO}{sinO} }{1- \frac{sinO}{cosO} } \\ =\frac{ \frac{sinO}{cosO} }{ \frac{sinO-cosO}{sinO} } +\frac{ \frac{cosO}{sinO} }{ \frac{cosO-sinO}{cosO} } \\ = \frac{sin^2O}{cosO(sinO-cosO)} + \frac{cos^2O}{sinO(cosO-sinO)} \\ = \frac{sin^2O}{cosO(sinO-cosO)} - \frac{cos^2O}{sinO(sinO-cosO)} \\ = \frac{sin^3O-cos^3O}{sinOcosO(sinO-cosO)} \\ =\frac{(sinO-cosO)(sin^2O+cos^2O+sinOcosO)}{sinOcosO(sinO-cosO)} \\ = \frac{1+sinOcosO}{sinOcosO} \\ \end{gathered}$

= $[1 +(1/cosecOsecO)]/[1/cosecOsecO]$

=[(cosecOsecO+1)/cosecOsecO]*[cosecOsecO]

=cosecOsecO+1

=RHS

Hence proved *☑

Answer 2

Answer:

Refer attached image for your answer.