Question

If $\sec \alpha =\frac{5}{4}$ verify that $\frac{\tan \alpha }{1+{\tan }^2\alpha }=\frac{\sin \alpha }{\sec \alpha }$

Answer 1

We have ,

$sec\alpha =\frac{Hypotenuse}{base}=\frac{5}{4}$

so , we draw a right triangle ABC , right angled at B such that

$\angle$ BAC = $\alpha$ , base = AB = 4 and Hypotenuse = AC = 5

By pyathagoras theorem , we have

$A{C}^2=A{B}^2+B{C}^2$

$\Rightarrow 5^2=4^2+B{C}^2$

$\Rightarrow =B{C}^2=25-16=9$

$\Rightarrow$ BC = 3

$\therefore tan\alpha =\frac{BC}{AB}=\frac{3}{4}andsin\alpha =\frac{BC}{AC}=\frac{3}{5}$

now $\frac{tan\alpha }{1+ta{n}^2\alpha }=\frac{\frac{3}{4}}{1+{\left(\frac{3}{4}\right)}^2}=\frac{\frac{3}{4}}{1+\frac{9}{16}}=\frac{\frac{3}{4}}{\frac{16+9}{16}}=\frac{\frac{3}{4}}{\frac{25}{16}}=\frac{3}{4}\times \frac{16}{25}=\frac{12}{25}$

and $\frac{sin\alpha }{sec\alpha }=\frac{3/5}{5/4}=\frac{3}{5}\times \frac{4}{5}=\frac{12}{15}$

$\therefore \frac{tan\alpha }{1+ta{n}^2\alpha }=\frac{sin\alpha }{sec\alpha }$