If tanA = 1 / √x(x^2+x+1), tanB = √x / √x^2+x+1 and tanC = (x^−3+x^−2+x^−1)^1/2, 0 < A, B, C < π/2,

Question

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

(1) \(\mathrm{C}\)

(2) \(\pi-C\)

(3) \(2 \pi-C\)

(4) \(\frac{\pi}{2}-C\)

Answer 1

Correct option is (1) \(\mathrm{C}\)

Finding \(\tan (A+B)\) we get

\(\Rightarrow \tan (A+B)=\)

\(\frac{\tan A+\tan B}{1-\tan A \tan B}=\cfrac{\frac{1}{\sqrt{x\left(x^{2}+x+1\right)}}+\frac{\sqrt{x}}{\sqrt{x^{2}+x+1}}}{1-\frac{1}{x^{2}+x+1}}\)

\(\Rightarrow \tan (\mathrm{A}+\mathrm{B})=\frac{(1+x)\left(\sqrt{x^{2}+x+1}\right)}{\left(x^{2}+x\right)(\sqrt{x})}\)

\(\tan (A+B)=\frac{\sqrt{x^{2}+x+1}}{x \sqrt{x}}=\tan C\)

\(A+B=C\)