Question

Given cos theta = - 2/5 and sin theta < 0, find the six trigonometric values. I need help with this

Answer 1

`\sin\,\theta =-(√(21) )/(5)`

`\tan\,\theta =(√(21) )/(2)`

`\sec\,\theta = (-5)/(2)`

`cosec\,\theta =(-5)/(√(21) )`

`\cot\,\theta =(2)/(√(21) )`

Explanation:

`\cos\theta =(-2)/(5)<0\\sin\theta <0`

As both
`sin\,\theta<0,\,\cos\,\theta <0`,

`\theta` lies in the third quadrant.

In the third quadrant,

`\sin\theta<0,\,\cos\,\theta <0\,,\,\sec\,\theta<0,\,cosec\theta<0,\,tan\,\theta>0,\,cot\,\theta>0`

`\sin\,\theta =-√(1-\cos^2\,\theta) \\=-\sqrt{1-((-2)/(5))^2 } \\\\=-\sqrt{1-(4)/(25) }\\\\=-\sqrt{(25-4)/(25) }\\\\=-(√(21) )/(5)`

`\tan\,\theta = (\sin\,\theta)/(\cos\,\theta )\\\\=((-√(21) )/(5) )/((-2)/(5) )\\\\=(√(21) )/(2)`

`\sec\,\theta =(1)/(\cos\,\theta )\\\\=(1)/((-2)/(5) )\\\\=(-5)/(2)`

`\ cosec \,\theta = (1)/(sin\,\theta )\\\\=(1)/((-√(21) )/(5) )\\\\=(-5)/(√(21) )`

`\cot\,\theta =(1)/(\tan\,\theta)\\\\=(1)/((√(21) )/(2) )\\\\=(2)/(√(21) )`