Question

in a right angled triangle ABC right angle at b and 15cosA=8sin A find cosec A -sec A dvided by Cose A+seca​

Answer 1

this is answer ..........

Answer 2

Answer:

-0.3

Step-by-step explanation:

Given,

$15CosA=8SinA$

$\frac{SinA}{CosA}=\frac{15}{8}$

$TanA=\frac{15}{8}$

BY PYTHAGORAS THEOREM,

$CosecA=\frac{17}{15}$

$SecA=\frac{17}{8}$

REQUIRED TO FIND:

$\frac{CosecA-SecA}{cosecA+SecA}$

First let's find;

$CosecA-SecA=\frac{17}{15}-\frac{17}{8}$

$CosecA-SecA=\frac{-119}{120}$

$CosecA+SecA=\frac{17}{15}+\frac{17}{8}$

$CosecA+SecA=\frac{391}{120}$

NOW,

$\frac{CosecA-SecA}{cosecA+SecA}$

$=\frac{\frac{-119}{120}}{\frac{391}{120}}$

$=\frac{-119}{391}$

OR

$=-0.3$