How do you evaluate $tan(pi/2+pi/6) . cot (pi-pi/6) . cos (pi/2-pi/6)$ ?

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Answer 1 · 2016-08-13T05:01:22

Using identities for tan(A+B) and cos(A+B)

So tan(A-B)= $"tanA-tanB"/"(1+tanAtanB)"$
So cot(A-B) will be the reciprocal - $"(1+tanAtanB)"/"tanA-tanB"$
So A- $pi$ and B is $-pi/6$
So =

"(tan( $2pi/3$ )" x "(cos( $pi/3$ ))"x ("(1+tan( $pi)tan($ -pi/6 $))")/"tan($ pi $)-tan($ -pi/6 $)"$

= $-sqrt3$ x $1/2$ x $(\frac{1}{\sqrt{3}}0+1)$ / $0-\frac{1}{\sqrt{3}}$

Solving will give 1.5

How do you evaluate tan(pi/2+pi/6) . cot (pi-pi/6) . cos (pi/2-pi/6)tan(pi/2+pi/6) . cot (pi-pi/6) . cos (pi/2-pi/6)?