Question

Given that tan (A+B)= tanA + tanB/ 1-tanA.tanB , find the value of tan 75 degree and tan 90 degree by taking suitable values of A and B. 

Answer 1

$tan(75) = tan(30+45) \\ tan(30+45)= \frac{tan30+tan45}{1-tan30.tan45} \\ = \frac{( \frac{1}{ \sqrt{3} }+1) }{1- \frac{1}{ \sqrt{3} }.1 } \\ = \frac{ \frac{1+ \sqrt{3} }{ \sqrt{3} } }{ \frac{ \sqrt{3} -1}{ \sqrt{3} } } \\tan75 = \frac{ \sqrt{3}+1 }{ \sqrt{3}-1 }$
Now, Tan(90)= tan(45+45)
$tan(45+45) = \frac{tan45+tan45}{1-tan45.tan45} \\ = \frac{1+1}{1-1*1} = \frac{1}{0}$ 
= infinity

Answer 2

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