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Question

Find the value of 1+sin2A+1sin2A1+sin2A1sin2A when |tanA|<1 and |A| is acute

A
tanA
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B
tanA
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C
cotA
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D
cotA
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Solution

The correct option is C. cotA

Given expression is 1+sin2A+1sin2A1+sin2A1sin2A

=sin2A+cos2A+sin2A+sin2A+cos2Asin2A1+sin2A1sin2A


=(cosA+sinA)2+(cosAsinA)2(cosA+sinA)2(cosAsinA)2

=|cosA+sinA|+|cosAsinA||cosA+sinA||cosAsinA|

=cosA+sinA+cosAsinAcosA+sinA(cosAsinA)

(π4<A<π4 and in this interval cosA>sinA)

=2cosA2sinA=cotA when |tanA|<1 and |A| is acute.

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