sin(3π-A)cos(A-π/2)tan(3π/2-A)
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cosec(13π/2+A)sec(3π+A)cot(A-π/2)=cos^4A
Answers
Step-by-step explanation:
sin(3π - A) = sin{2π + ( π - A)}
= sin(π - A) = sinA
cos(A - π/2) = cos{-(π/2 - A)} = cos(π/2 - A)[as we know, cos(-x) = cosx]
= cos(π/2 - A) = sinA
tan(3π/2 - A) = cotA [ in 3rd quadrant, tan, cot are positive ]
cosec(13π/2 + A) = cosec(6π + π/2 + A)
= cosec(π/2 + A) = -secA
sec(3π + A) = sec(2π + π + A)
= sec(π + A) = -secA
cot(A - π/2) = -cot(π/2 - A) = -tanA
now,
LHS
{sin(3π - A)cos(A - π/2)tan(3π/2 - A)}{cosec(13π/2 + A)sec(3π + A)cot(A - π/2)}
= {sinA sinA cotA}/{-secA (-secA)(-tanA)}
= {sin²A cotA}/{sec²A tanA}
= {sin²A × cosA/sinA}/{1/cos²A × sinA/cosA}
= cosA/{1/cos³A}
= cos⁴A
RHS
Hope it helps!
Given : sin(3π-A)cos(A-π/2)tan(3π/2-A)÷cosec(13π/2+A)sec(3π+A)cot(A-π\2)
To Find : Simplify
Solution:
sin(3π-A) = Sin(2π+π-A) = Sin(π-A) = SinA
cos(A-π/2) = Cos(π/2 - A) = SinA
tan(3π/2-A) = CotA
cosec(13π/2+A) = cosec(π/2+A) = secA
sec(3π+A ) = sec(π+A ) = -secA
cot(A-π\2) = -tanA
SinA . SinA . CotA / ( ( secA )(-secA )(-tanA)
= SinA cosA / sec²AtanA
= CosAcosA. cos²A
= Cos⁴A
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