Question

Find sin 2x, cos 2x, tan 2x if secx = `(-13)/5, pi/2 < x < pi`

Answer 1

Since `pi/2 < x < pi, x` lies in the second quadrant.

We have sec x = `-13/5`

∴ tan²x = sec²x – 1 = `(-13/5)^2 - 1`

= `169/25 - 1`

= `144/25`

∴ tanx = `±12/5`

But x lies in the second quadrant

∴ tanx is negative

∴ tanx = `-12/5`

∴ sin2x = `(2tanx)/(1 + tan^2x)`

= `(2(-12/5))/(1 + (-12/5)^2)`

= `((-24/5))/(1 + 144/25)`

= `((-24)/5)/(169/25)`

= `(-24)/5 xx 25/169`

= `(-120)/169`

cos2x = `(1 - tan^2x)/(1 + tan^2x)`

= `(1 - (-12/5)^2)/(1 + (-12/5)^2`

= `(1 - 144/25)/(1 + 144/25)`

= `(25 - 144)/(25 + 144)`

= `-119/169`

tan2x = `(2tanx)/(1 - tan^2x)`

= `(2(-12/5))/(1 - (-12/5)^2`

= `((-24/5))/(1 - 144/25)`

= `((-24/5))/((-119/25)`

= `24/5 xx 25/119`

= `120/119`