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JEE Main
Exam
JEE Main
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

JEE Main · Mathematics · Trigonometric Ratios & Identities

If tanA = 1 x(x^ 2 + x + 1) , tanB = x x^ 2 + x + 1 and tanC = (x^...

If tanA=1x(x2+x+1),tanB=xx2+x+1tanA = \dfrac{1}{x\left(x^{2} + x + 1\right)}, tanB = \dfrac{x}{x^{2} + x + 1} and tanC=(x−3+x−2+x−1)12,0<A,B,C<π2tanC = \left(x^{- 3} + x^{- 2} + x^{- 1}\right)\dfrac{1}{2}, 0 < A, B, C < \dfrac{π}{2}, then A+BA + B is equal to:

Options

  1. ACC
  2. Bπ−C\pi - C
  3. C2π−C2\pi - C
  4. Dπ2−C\dfrac{π}{2} - C

Correct answer

A. CC

Step-by-step solution

Given, tanA=1x(x2+x+1),tanB=xx2+x+1tanA = \dfrac{1}{x\left(x^{2} + x + 1\right)}, tanB = \dfrac{x}{x^{2} + x + 1} and tanC=(x−3+x−2+x−1)12=1+x+x2xxtanC = \left(x^{- 3} + x^{- 2} + x^{- 1}\right)\dfrac{1}{2} = \dfrac{1 + x + x^{2}}{x\sqrt{x}}

We know that,

tan⁡(A+B)=tanA+tanB1−tanAtanB\tan \left(A + B\right) = \dfrac{tanA + tanB}{1 - tanAtanB}

⇒tan⁡(A+B)=1x(x2+x+1)+xx2+x+11−1x2+x+1⇒ \tan \left(A + B\right) = \dfrac{\dfrac{1}{x\left(x^{2} + x + 1\right)} + \dfrac{x}{x^{2} + x + 1}}{1 - \dfrac{1}{x^{2} + x + 1}}

⇒tan⁡(A+B)=(1+x)(x2+x+1)(x2+x)(x)⇒ \tan \left(A + B\right) = \dfrac{\left(1 + x\right)\left(\sqrt{x^{2} + x + 1}\right)}{\left(x^{2} + x\right)\left(\sqrt{x}\right)}

⇒tan⁡(A+B)=(1+x)(x2+x+1)(x2+x)(x)⇒ \tan \left(A + B\right) = \dfrac{\left(1 + x\right)\left(\sqrt{x^{2} + x + 1}\right)}{\left(x^{2} + x\right)\left(\sqrt{x}\right)}

⇒tan⁡(A+B)=x2+x+1xx⇒ \tan \left(A + B\right) = \dfrac{x^{2} + x + 1}{x\sqrt{x}}

⇒tan⁡(A+B)=tanC⇒ \tan \left(A + B\right) = tanC

⇒A+B=C⇒ A + B = C

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