Challenging Trigonometry Question (ACS(I) Sec 3)

maths tuition trigonometry


(a) \tan (\alpha + \beta )=\frac{10}{x}

Using the formula,

\displaystyle \frac{\tan \alpha + \tan \beta}{1-\tan \alpha \tan \beta}=\frac{10}{x}

\displaystyle \frac{\frac{5}{x}+\tan \beta}{1-(\frac{5}{x})\tan \beta}=\frac{10}{x}


5+x\tan\beta =10-\frac{50}{x}\tan \beta

(x+\frac{50}{x})\tan\beta =5

\displaystyle \tan \beta =\frac{5}{x+\frac{50}{x}}=\frac{5x}{x^2+50}

(b) The trick here is to break up 2\alpha +\beta into \alpha + (\alpha +\beta )

\displaystyle \begin{array}{rcl}    \tan (2\alpha +\beta )&=& \tan (\alpha + (\alpha + \beta ))\\    &=& \frac{\tan \alpha + \tan (\alpha + \beta )}{1-\tan \alpha \tan (\alpha + \beta )}\\    &=& \frac{ \frac{5}{x}+\frac{10}{x} }{ 1-(\frac{5}{x})(\frac{10}{x}) }\\    &=& \frac{\frac{15}{x}}{\frac{x^2-50}{x^2}}\\    &=& \frac{15x}{x^2-50}    \end{array}


Since 2\alpha +\beta is acute (1st quadrant), \tan (2\alpha +\beta ) is positive.

x^2-50 >0



Author: mathtuition88

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