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Q.

cosec2Acot2Asec2Atan2A(cot2Atan2A)(sec2Acosec2A1)=

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a

0

b

1

c

1

d

2

answer is A.

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Detailed Solution

Put A=45°

cosec245°=sec245°tan245°(cot245tan245°)

(sec245°cosce245°1)

=2(1)2(1)(11)  (2(2)1)

=0

1sin2Acos2Asin2A1cos2Asin2Acos2A(cosAsin2AsinAcos2A)(1cos2A1sin2A1)

cos2Asin4Asin2Acos4A(cos4sin4Asin2Acos2A)(1cos2Asin2Asin2Acos2A)

cos6Asin6Asin4Acos4Acos4Asin4Acos6Asin2A+cos2Asin4Asin2Acos2A

=(cos2A)3(sin2A)3sin4Acos4A[(cos2A)2(sin2A)2cos2Asin2A(cos4Asin4A)sin4Acos4A]

=(cos2Asin2A)(1sin2Acos2A)sin4Acos4A[(cos2Asin2A)cos2Asin2A(cos2Asin2A)sin4cos4A] =0

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