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11Alexandr11 [23.1K]
3 years ago
10

Question is attached.Step by step explanation needed!!​

Mathematics
2 answers:
gregori [183]3 years ago
4 0

Your answer is attached.

  • Hence Proved.

goldfiish [28.3K]3 years ago
3 0

LHS

\\ \sf\longmapsto \dfrac{tanA+secA-1}{tanA-secA+1}

As we know

\\ \boxed{\sf sec^2A-tan^2A=1}

\\ \sf\longmapsto \dfrac{tanA+secA-(sec^2A-tan^2A)}{tanA-secA+1}

\\ \sf\longmapsto \dfrac{tanA+secA\cancel{(1-secA+tanA)}}{\cancel{(1-secA+tanA)}}

\\ \sf\longmapsto tanA+secA

\\ \sf\longmapsto \dfrac{sinA}{cosA}+\dfrac{1}{cosA}

\\ \sf\longmapsto \dfrac{1+sinA}{cosA}

\\ \sf\longmapsto RHS

Hence proved

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