Question

$\Delta ABC$ is right angled at B and $5\times sinA=3$. Find cos C + tan A + cosec C.

Answer 1

Consider $\Delta ABC$ is right angled at B.

$5\times sinA=3$ (given)

$\Rightarrow sinA=\frac{3}{5}=\frac{BC}{AC}$

So, if BC = 3k, then AC = 5k where k is a positive number.

Applying Pythagoras' theorem,

$A{B}^2+B{C}^2=A{C}^{2}A{B}^2=(5k{)}^2-(3k{)}^2=16k^{2}AB=4k$

$cosC=\frac{BC}{AC}=\frac{3}{5}$

$tanA=\frac{BC}{AB}=\frac{3}{4}$

$cosecC=\frac{AC}{AB}=\frac{5}{4}$

$\Rightarrow cosC+tanA+cosecC$

$=\frac{3}{5}+\frac{3}{4}+\frac{5}{4}$

$=\frac{12+15+25}{20}=\frac{13}{5}$