Question

Please solve this question Step By Step Explanation. ​

Answer 1

Step-by-step explanation:

$\color{purple} \text{ \boxed{ \tt \: Solution :} \qquad \: \qquad \: We know that} \\ \\ \color{red}\[ \tan (\propto+\beta)=\frac{\tan \propto+\tan \beta}{1-\tan \propto \tan \beta} \]$

$\tt \color {blue}\therefore \tan (\alpha+\beta)= \left[\begin{array}{l} \tt\dfrac{\dfrac{1}{\sqrt{x\left(x^{2}+x+1\right)}}+\dfrac{\sqrt{x}}{\sqrt{x^{2}+x+1}}} { 1-\dfrac{1}{\sqrt{x\left(x^{2}+x+1\right)}} \dfrac{\sqrt{x}}{\sqrt{x^{2}+x+1}}}\end{array}\right]$

$\begin{array}{ll} \tt \color{darkorange} \\ = & \tt \color{darkorange}\dfrac{(x+1) \sqrt{x^{2}+x+1}}{\sqrt{x} x(x+1)} \\ \\ \tt \color{red} = & \tt \color{red} \sqrt{\dfrac{x^{2}}{x^{3}}+\dfrac{x}{x^{3}}+\dfrac{1}{x^{3}}} \\ \\ \tt \color{maroon}= & \tt \color{maroon}\sqrt{x^{-1}+x^{-2}+x^{-3}} \\ \\ \tt \color{navy} = & \tt \color{navy}\tan \gamma \\ \\ \boxed{ \tt \color{darkgreen}\therefore \alpha+\beta=\gamma} & \end{array}$