Question

Given tan x= 1/3 and cos x < 0 ,find the value of csc x

Answer 1

`\tan x=\frac13=(\sin x)/(\cos x)`


`\tan x>0`, but we're told that
`\cos x<0`, which means
`\sin x<0`. So we should expect
`\csc x<0` as well.

Recall the Pythagorean identity:


`\sin^2x+\cos^2x=1\implies\tan^2x+1=\sec^2x\implies\sec x=\pm√(\tan^2x+1)`

But we know
`\cos x<0`, so
`\sec x<0`, too.


`\implies\sec x=-√(\tan^2x+1)=-√(\frac19+1)=-\sqrt{\frac{10}9}`

`\implies\cos x=-\sqrt{\frac9{10}}`


`\sin^2x+\cos^2x=1\implies\sin x=-\sqrt{1-\left(-\sqrt{\frac9{10}}\right)^2}=-\sqrt{\frac1{10}}`

`\implies\csc x=-√(10)`