Question

The question is in the screenshot

Answer 1

Answer:

b. -36/77

Step-by-step explanation:

As 0 < x < $\pi$/2 => tan x > 0

As 0< y <  $\pi$/2 => tan y > 0

We have the formula:

• $\frac{1}{sin^{2}x } =1 + cot^{2}x$ => $\frac{1}{(8/17)^{2} }$ = $1+cot^{2}x$ => $cot^{2} x=225/64$

As tan x = 1/(cotx) => $tan^{2}x = \frac{1}{cot^{2}x} =\frac{1}{225/64} = \frac{64}{225}$

As tan x > 0 => tan x = 8/15

• $\frac{1}{cos^{2}y } = 1+tan^{2}y$ => $\frac{1}{(3/5)^{2} } = 1 + tan^{2} y$ => $tan^{2}y=\frac{16}{9}$

As tan y > 0 => tan y = 4/3

As tan (x-y)= $\frac{tan x - tan y}{1+ tanx * tany}$ $=\frac{(8/15)-(4/3)}{1+(8/15)*(4/3)} = \frac{-36}{77}$