If tan ⁡A= 1 x(x^2+x+1) , tan ⁡B= x x^2+x+1 and tan ⁡C=(x^-3+x^-2+x^-1)^ 12 ,0

Question

If tan ⁡A= 1 x(x^2+x+1) , tan ⁡B= x x^2+x+1 and tan ⁡C=(x^-3+x^-2+x^-1)^ 12 ,0<A ,B ,C< 2, then A+B is equal to :

Accepted Answer

Correct answer: A. (a) Finding tan ⁡ (A+B ) we get ⇒ tan ⁡ (A+B )= tan ⁡A+ tan ⁡B1- tan ⁡ Atan ⁡B= 1 x(x^2+x+1)+ x x^2+x+11- 1x^2+x+1 ⇒ tan ⁡ (A+B )= (1+x ) ( x^2+x+1 )(x^2+x) ( x ) (1+x ) ( x^2+x+1 )(x^2+x) ( x ) tan ⁡ (A+B )= x^2+x+1x x= tan ⁡C A+B=C

$If tan A = \frac{1}{x ( x ^{2} + x + 1 )}, tan B = \frac{x}{x ^{2} + x + 1} and tan C = (x^{- 3} + x^{- 2} + x^{- 1})^{\frac{1}{2}}, 0 < A, B, C < \frac{π}{2}, then A + B is equal to :$

Solution

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