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Q.

cosec2Acot2A−sec2Atan2A−(Cot2A−tan2A)(sec2Acosec2A−1)=

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a

0

b

1

c

2

d

3

answer is A.

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Detailed Solution

cos2Asin4A-sin2Acos4A-cos2Asin2A-sin2Acos2A1sin2Acos2A-1 cos6A-sin6A-cos4A-sin4A1-sin2Acos2Asin4Acos4A but cos6A-sin6A=cos2A3-sin2A3=cos2A-sin2Acos4A+sin4A+cos2Asin2A cos4A+sin4A=1-2sin2Acos2A cos4A-sin4A=cos2A+sin2Acos2A-sin2A
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cosec2Acot2A−sec2Atan2A−(Cot2A−tan2A)(sec2Acosec2A−1)=