Answer :

cecepham

Answer:

b. -36/77

Step-by-step explanation:

As 0 < x < [tex]\pi[/tex]/2 => tan x > 0

As 0< y <  [tex]\pi[/tex]/2 => tan y > 0

We have the formula:

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

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

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

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

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

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

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