How do you evaluate $tan (\frac{π}{2} + \frac{π}{6}) . cot (π - \frac{π}{6}) . cos (\frac{π}{2} - \frac{π}{6})$ $tan(pi/2+pi/6) . cot (pi-pi/6) . cos (pi/2-pi/6)$ ?

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How do you evaluate $tan (\frac{π}{2} + \frac{π}{6}) . cot (π - \frac{π}{6}) . cos (\frac{π}{2} - \frac{π}{6})$ $tan(pi/2+pi/6) . cot (pi-pi/6) . cos (pi/2-pi/6)$ ?

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Answer 1 · 2016-08-13T05:01:22

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

Explanation:

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

"(tan( $2 \frac{π}{3}$ $2pi/3$ )" x "(cos( $\frac{π}{3}$ $pi/3$ ))"x ("(1+tan( $π) tan ($ $pi)tan($ -pi/6 $)))/ tan ($ $))")/"tan($ pi $) - tan ($ $)-tan($ -pi/6 $)$ $)"$

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

Solving will give 1.5

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How do you evaluate $tan (\frac{π}{2} + \frac{π}{6}) . cot (π - \frac{π}{6}) . cos (\frac{π}{2} - \frac{π}{6})$ $tan(pi/2+pi/6) . cot (pi-pi/6) . cos (pi/2-pi/6)$ ?

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Explanation:

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How do you evaluate $tan (\frac{π}{2} + \frac{π}{6}) . cot (π - \frac{π}{6}) . cos (\frac{π}{2} - \frac{π}{6})$ $tan(pi/2+pi/6) . cot (pi-pi/6) . cos (pi/2-pi/6)$ ?