Mathematics - Trigonometric Ratios & Identities Question with Solution | TestHub

Given,$\tan A=\frac{1}{\sqrt{x\left(x^2+x+1\right)}},\tan B=\frac{\sqrt{x}}{\sqrt{x^2+x+1}}$and$\tan C={\left(x^{-3}+x^{-2}+x^{-1}\right)}^{\frac{1}{2}}=\frac{\sqrt{1+x+x^2}}{x\sqrt{x}}$

We know that,

$\tan \left(A+B\right)=\frac{\tan A+\tan B}{1-\tan A\tan B}$

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

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

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

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

$\Rightarrow \tan \left(A+B\right)=\tan C$

$\Rightarrow A+B=C$