cosA/1+sin A +1+sinA/cosA =1/secA -tanA
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i guess question is wrong...we can prove
cosA/(1+sinA)+(1+sinA)/cosA=2 secA
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cosA/1+sinA + 1+sinA/cosA=
=>cos²A+(1+sinA)²/cosA(1+sinA)
=>(cos²A+sin²A+1+2sinA)/cosA(1+sinA)
=>2[1+sinA]/cos(1+sinA)
=>2/cosA=>2secA
cosA/(1+sinA)+(1+sinA)/cosA=2 secA
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cosA/1+sinA + 1+sinA/cosA=
=>cos²A+(1+sinA)²/cosA(1+sinA)
=>(cos²A+sin²A+1+2sinA)/cosA(1+sinA)
=>2[1+sinA]/cos(1+sinA)
=>2/cosA=>2secA
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