Question

In a right angled triangle ABC right angled at B, $5\times sinA=3$.
Find the value of $cosC+tanA+cosecC.$

• A. $\frac{12}{5}$
• B. $\frac{13}{5}$
• C. $\frac{4}{5}$
• D. $\frac{13}{12}$

Answer 1

The correct option is B $\frac{13}{5}$


Given, triangle ABC is right angled at B.
$\text{Also,}5\times sinA=3$

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

Let BC = 3k, AC = 5k $(\because BC:AC=3:5)$

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

$cosC=\frac{Adjacentside}{Hypotenuse}=\frac{BC}{AC}=\frac{3}{5}tanA=\frac{OppositeSide}{Hypotenuse}=\frac{BC}{AB}=\frac{3}{4}cosecC=\frac{Hypotenuse}{Oppositeside}=\frac{AC}{AB}=\frac{5}{4}$

Hence,
$cosC+tanA+cosecC=\frac{3}{5}+\frac{3}{4}+\frac{5}{4}=\frac{12+15+25}{20}=\frac{13}{5}$