Question

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

• A. $\frac{12}{5}$
• B. $\frac{13}{12}$
• C. $\frac{13}{5}$
• D. $\frac{12}{5}$

Answer 1

The correct option is C

$\frac{13}{5}$

$\Delta$ ABC is right angled at B,
Given,$5\times sinA=3sinA=\frac{3}{5}=\frac{BC}{AC}$
Which means BC=3k,AC=5k since, BC:AC=3:5
Applying Pythagoras theorem, $A{B}^2+B{C}^2=A{C}^{2}A{B}^2=(5k{)}^2-(3k{)}^2=16k^{2}AB=4kcosC=\frac{BC}{AC}=\frac{3}{5}tanA=\frac{BC}{AB}=\frac{3}{4}cosecC=\frac{AC}{AB}=\frac{5}{4}cosC+tanA+cosecC=\frac{3}{5}+\frac{3}{4}+\frac{5}{4}=\frac{12+15+25}{20}=\frac{13}{5}$