Question

If tanΘ= -4/3 and Θ lies in the second quadrant, then sinΘ-2cosΘ=?

Answer 1

sinΘ-2cosΘ = (sin Θ/ cos Θ) -(2cos Θ/cosΘ) = tan Θ - 2 = -4/3 - 2 = ???

Answer 2

why we need to divided by cosΘ??

Answer 3

you can't just divide by cos theta, you need to multiply also.

Answer 4

i would suggest you find sec theta first, from the identity
sec^2 theta = 1+ tan^2 theta
and then, cos = 1/ sec,
sin = tan * cos.

Answer 5

i have learnt sec yet, i have learnt sin cos tan only so far

Answer 6

there is a picture of an angle whose tangent is \(\frac{4}{3}\)

Answer 7

you need the third side, which you get via pythagoras or by remembering the 3 - 4 - 5 right triangle

Answer 8

and?

Answer 9

now you see that \(\sin(\theta)=\frac{4}{5}\) and \(\cos(\theta)=\frac{3}{5}\) except that since you are in quadrant II you have \(\cos(\theta)=-\frac{3}{5}\) because in quadrant II cosine is negative

Answer 10

you want
\[\sin(\theta)-2\cos(\theta)\] and you know the numbers that you need
\[\frac{4}{5}-2\times (-\frac{3}{5})=\frac{4}{5}+\frac{6}{5}\]

Answer 11

rather easy if you draw a triangle
otherwise a pita

Answer 12

2?

Answer 13

Alternate way to look at the same thing,

so the angle inside triangle is \(\pi - \theta\)
so, \(\cos(\pi-\theta)=3/5, -\cos \theta=3/5 , \cos \theta = -3/5\)