Question

Let 0 be an angle in quadrant

Answer 1

If angle theta is given as;

`\cos \theta=-(4)/(5)`

And angle theta is in the third quadrant, then we can find the value of y by using the pythagoras' theorem as follows;

`\begin{gathered} r^2=x^2+y^2 \\ When\text{ cos}\theta=-(4)/(5),\text{ and cos}\theta=(x)/(r) \\ \text{Then,} \\ x=-4,r=5 \\ \text{Therefore,} \\ 5^2=-4^2+y^2 \\ 25=16+y^2 \\ \text{Subtract 16 from both sides;} \\ 9=y^2 \\ \text{Take the square root of both sides;} \\ y=3 \end{gathered}`

Note also that;

`\sin \theta=(y)/(r)`

However, the angle theta is in the third quadrant where y is also negative. Therefore, sin theta would be;

`\begin{gathered} \sin \theta=(y)/(r) \\ \sin \theta=-(3)/(5) \end{gathered}`

Note also that cosec theta is given as;

`\begin{gathered} \csc \theta=(1)/(\sin \theta) \\ \csc \theta=(1)/(-(3)/(5)) \\ \csc \theta=-(5)/(3) \end{gathered}`

Note also that, tan theta is given as;

`\tan \theta=(\sin \theta)/(\cos \theta)`

Therefore, we would have;

`\begin{gathered} \tan \theta=-(3)/(5)\text{ / -}(4)/(5) \\ \tan \theta=-(3)/(5)*-(5)/(4) \\ \tan \theta=(3)/(4) \end{gathered}`

ANSWER: