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Prove that (tan x+ sinx)/(tanx-sinx)=(secx+1)/(secx-1)

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Solution: We have ,

LHS $\= (tanx+sinx)/(tanx- sinx)\ \Rightarrow (sinx/cosx +sinx)/(sinx/cosx-sinx)\ \Rightarrow [sinx(1/cosx+1)]/[sinx(1/cosx -1)]\ \ Rightarrow (secx+1)/(secx-1)$

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Deependra Verma

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