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​prove that secB-cosecB/secB+cosecB=tanB-1/tanB+1with all the steps

​prove that secB-cosecB/secB+cosecB=tanB-1/tanB+1with all the steps

Grade:12

1 Answers

Akshay
15 Points
7 years ago
RHS = (tanB - 1)/(tanB + 1) = (tanB - tanπ/4)/(1 + tanBtanπ/4) = tan(B - π/4) = sin(B - π/4)/cos(B - π/4) = (sinBcosπ/4 - cosBsinπ/4)/(cosBcosπ/4 + sinBsinπ/4) = (sinB - cosB)/(cosB + sinB) = (1/cosecB - 1/secB)/(1/secB + 1/cosecB) = [(secB - cosecB)/(cosecBsecB)]/[(cosecB + secB)/(cosecBsecB)] = (secB - cosecB)/(secB + cosecB) . RHS = LHS hence proved . Please approve my answer if you like it .

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