Question

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Answer 1

Step-by-step explanation:

Given,

$\frac{Cos20^{o}+Sin20^o }{Cos20^o-Sin20^o} = Tan65^o$

L.H.S,

$=\frac{Cos20^o[1+\frac{Sin20^o}{Cos20^o}]}{Cos20^o[1-\frac{Sin20^o}{Cos20^o}]} \\=\frac{1+Tan20^o}{1-Tan20^o}$

$=\frac{Tan45^o+Tan20^o}{Tan45^o-Tan20^o}$ ∴[$Tan45^o=1$]

$=\frac{Tan45^o+Tan20^o}{1-Tan45^o.Tan20^o}\\=Tan(45+20)^o$∴[$\frac{TanA+TanB}{1-TanA.TanB}=Tan(A+B)$]

$=Tan65^o$

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Answer 2

Step-by-step explanation:

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