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Question

# 1+cosA−sinA1+cosA+sinA=secA−tanA=1−sinAcosA=√1−sinA1+sinA

A
True
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B
False
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C
Ambiguous
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D
Data insufficient
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Solution

## The correct option is B True1+cosA−sinA1+cosA+sinA=(1+cosA−sinA)2(1+cosA)2−sin2A=1+cos2A+sin2A−2cosAsinA−2sinA+2cosA(1+cosA)2−(1−cosA)(1+cosA)=2−2cosAsinA−2sinA+2cosA(1+cosA)(1+cosA−1+cosA)=2(1+cosAsinA−sinA+cosA)2cosA(1+cosA)=1+cosAsinA−sinA+cosAcosA(1+cosA)=(1+cosA)−sinA(1+cosA)cosA(1+cosA)=secA−tanA=1−sinAcosA=√(1−sinA)2cos2A=√(1−sinA)21−sin2A=√(1−sinA)1+sinA

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