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Question

If tanx=34,π<x<3π2, find the value of sinx2,cosx2,tanx2

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Solution

tanx=34

tan2x=2tanx1tan2x

tan2(x2)=34

2tanx/21tan2x/2=34

tanx/2=t

2t1t2=34

8t=33t2

3t2+8t3=0

3t29tt3=0

3t(t+3)1(t+3)=0

t=13 t=3

tanx/2=13 or tanx/2=3

since x2(π2, 3π4)

π2=2nd quadrant

tanx/2= negative

tanx/2=3

=sinx/2=310

cosx/2=110

tanx/2=3


1365326_1178271_ans_dbbfcefe3f9044979d42fb3c26e5fed1.png

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