Question

sin[2π-8]cos[π+A]tan[π/2-A]÷sin[π/2+A]cot[2π+A]sin[π-A]=​

Answer 1

Answer:

Given:

\frac{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)cot(2π+A)sin(π−A)sin(2π−A)cos(π+A)tan(2π−A)

To find:

The Simplified form of the given term

Solution:

We have given the ratio of the trigonometry where we just have to find the simplified form of the particular ratios first to get the value of the given term

we have the equation as

\frac{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)cot(2π+A)sin(π−A)sin(2π−A)cos(π+A)tan(2π−A)

now by the concept of the quadrant of the ratios we get

\frac{-sinA(-cosA)cotA}{cosAcotAsinA}cosAcotAsinA−sinA(−cosA)cotA

\frac{sinAcosAcotA}{cosAcotAsinA}cosAcotAsinAsinAcosAcotA

as the numerator and denominator of the fraction is equal so the fraction will give you answer as 1

Hence the simplified form will be 1