Question

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
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Answer 1

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!

Answer 2

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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