Question

1. If tan θ= -8/15 and if the terminal point determined by θ is in quadrant II, find cscθ sinθ+ tan θ.

Answer 1

Since

$\csc(x)=\dfrac{1}{\sin(x)}$

we have

$\csc(x)\sin(x)+\tan(x) = \dfrac{1}{\sin(x)}\cdot \sin(x)+\tan(x)=1+\tan(x)$

And since

$\tan(x)=\dfrac{8}{15}$

we have

$1+\tan(x) = 1-\dfrac{8}{15}=\dfrac{15}{15}-\dfrac{8}{15}=\dfrac{7}{15}$

Answer 2

The value of cscθ sinθ + tan θ is 7/15.

What is the relationship between cscθ and sinθ?

cscθ is the reciprocal of sinθ.

We can find the value of the given expression below:

The expression is given by cscθ sinθ + tan θ.

We can simplify this expression before solving it.

Let the expression be y = cscθ sinθ + tan θ.

cscθ = 1/sinθ

y = sinθ * 1/sinθ + tan θ

y = 1 + tan θ

The value of tan θ is given as -8/15.

We can substitute this above:

y = 1 + tan θ

= 1 - 8/15

= (15-8)/15

= 7/15

The value of the expression cscθ sinθ+ tan θ is found to be 7/15.

Therefore, the value of cscθ sinθ + tan θ is 7/15.

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