Question

If $\cos \text{A =}\frac{3}{5}\text{and}\text{sin}\text{B =}\frac{5}{13},$ then the value of $\frac{\tan \text{A}+\tan \text{B}}{1-\tan \text{A}\tan \text{B}}$ is

Answer 1

Solution :-

Given $cosA=\frac{Base}{Hypotenuse}=\frac{3}{5}$

$\Rightarrow$ perpendicular $=\sqrt{(3{)}^2-(3{)}^2}=4$

$\therefore tanA=\frac{4}{3}$

Similarly $sinB=\frac{5}{13}$

$\therefore tanB=\frac{5}{12}$

$\therefore \frac{tanA+tanB}{1-tanAtanB}=\frac{\frac{4}{3}+\frac{5}{12}}{1-\frac{4}{3}\times \frac{5}{12}}$

$=\frac{\frac{16+5}{12}}{\frac{36-20}{36}}=\frac{21\times 3}{16}$

$=\frac{63}{16}$