Question

How do you evaluate `cos^2(pi/3)-cot(pi/4)+sin(pi/6 )`?

Answer 1

`=1/4`

Explanation:

To evaluate `cos^2(pi/3)-cot(pi/4)+sin(pi/6 )`

As we know from table `cos(pi/3)=sqrt3/2,cot(pi/4)=1 &sin(pi/6 )=1/2`

`cos^2(pi/3)-cot(pi/4)+sin(pi/6 )`
`=>(sqrt3/2)^2-1+1/2`
`=>3/4-1+1/2=(3-4+2)/4=1/4`