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

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Answer 1 · 2016-05-07T11:42:08

$=1/4$

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

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

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

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