Question

If cos(α + β) = 3/5, sin(α - β)
= 5/13
and 0 < α,
β < π/4, then tan(2α) is equal to:
(A) 63/52
(B) 33/52
(C) 63/16
(D) 21/16

Answer 1

$\textbf{Given:}$

$cos(\alpha+\beta)=\frac{3}{5}$

$sin(\alpha-\beta)=\frac{5}{13}$

$cos(\alpha+\beta)=\frac{3}{5}$

$\text{Taking reciprocals}$

$sec(\alpha+\beta)=\frac{5}{3}$

$tan^2(\alpha+\beta)=sec^2(\alpha+\beta)-1$

$tan^2(\alpha+\beta)=\frac{25}{9}-1$

$tan^2(\alpha+\beta)=\frac{16}{9}$

$\implies\bf\,tan(\alpha+\beta)=\frac{4}{3}$

$sin(\alpha-\beta)=\frac{5}{13}$

$cosec(\alpha-\beta)=\frac{13}{5}$

$cot^2(\alpha-\beta)=cosec^2(\alpha-\beta)-1$

$cot^2(\alpha-\beta)=\frac{169}{25}-1$

$cot^2(\alpha-\beta)=\frac{144}{25}$

$cot(\alpha-\beta)=\frac{12}{5}$

$tan(\alpha-\beta)=\frac{5}{12}$

$\text{Consider,}$

$tan2\alpha$

$=tan[(\alpha+\beta)+(\alpha-\beta)]$

$=\displaystyle\frac{tan(\alpha+\beta)+tan(\alpha-\beta)}{1-tan(\alpha+\beta)\,tan(\alpha-\beta)}$

$=\displaystyle\frac{\frac{4}{3}+\frac{5}{12}}{1-(\frac{4}{3})(\frac{5}{12})}$

$=\displaystyle\frac{\frac{12+5}{12}}{1-(\frac{1}{3})(\frac{5}{3})}$

$=\displaystyle\frac{\frac{21}{12}}{1-\frac{5}{9}}$

$=\displaystyle\frac{\frac{21}{12}}{\frac{4}{9}}$

$=\displaystyle\frac{21}{12}{\times}\frac{9}{4}$

$=\displaystyle\frac{21}{4}{\times}\frac{3}{4}$

$=\displaystyle\frac{63}{16}$

$\therefore\boxed{\bf\tan\,2\alpha=\frac{63}{16}}$

$\implies\text{Option (C) is correct}$