Given that sin α = 3 / 5 and α is the second quadrant angle, then SiN4 α / [4sin & # 178; (π / 4 + α) Tan (π / 4 - α)] is equal to

Given that sin α = 3 / 5 and α is the second quadrant angle, then SiN4 α / [4sin & # 178; (π / 4 + α) Tan (π / 4 - α)] is equal to

sin4a/[4sin²(π/4+a)tan(π/4-a)]
=sin4a/[4cos²(π/4-a)tan(π/4-a)]
=sin4a/[4cos(π/4-a)sin(π/4-a)]
=sin4a/[2sin(π/2-2a)]
=sin4a/[2cos2a]
=sin2a
=2sinacosa
∵ α is the second quadrant angle
sina=3/5
∴cosa=-4/5
The original form
=2*3/5*(-4/5)
=-24/25
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