Question

Sin(2 兀-A)cos(兀+A)Tan(兀/2-A)/sin (兀/2+A)cot(2 兀+A)sin(兀-A)=

Answer 1

Given :

$\dfrac{\sin(2\pi-A)\cos(\pi+A)\tan(\frac{\pi}{2}-A)}{\sin(\frac{\pi}{2}+A)\cot(2\pi+A)\sin(\pi-A)}$

To Find : Evaluate

Solution:

$\dfrac{\sin(2\pi-A)\cos(\pi+A)\tan(\frac{\pi}{2}-A)}{\sin(\frac{\pi}{2}+A)\cot(2\pi+A)\sin(\pi-A)}$

sin(2π - A) =  -sinA

cos(π+A) =  -cosA

tan(π/2 - A)  = cotA  

Numerator  = -sinA. (-cosA)cotA   = sinAcosAcotA

sin(π/2 + A)  = cosA

cot(2π + A) = cotA

sin(π - A)  = SinA

Denominator = cosAcotAsinA = sinAcosAcotA

Numerator  = Denominator

=> Numerator  /  Denominator = 1

$\dfrac{\sin(2\pi-A)\cos(\pi+A)\tan(\frac{\pi}{2}-A)}{\sin(\frac{\pi}{2}+A)\cot(2\pi+A)\sin(\pi-A)}=1$

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