$\text{Given that,}\\\\\cos \theta = - \dfrac 23\\\\\implies \cos^2 \theta = \dfrac 49\\\\\implies 1- \sin^2 \theta = \dfrac 49\\\\\implies \sin^2 \theta =1-\dfrac 49\\\\\implies \sin^2 \theta =\dfrac 59\\\\\\\text{Now,}\\\\\cot^2 \theta = \dfrac{\cos^2 \theta}{\sin^2 \theta} \\\\\\\implies \cot^2 \theta = \dfrac{\tfrac 49}{\tfrac 59}\\\\\\\implies \cot^2 \theta = \dfrac 49 \times \dfrac 95\\\\\\\implies \cot^2 \theta = \dfrac 45\\\\\\$

$\implies \cot \theta = \pm\sqrt{\dfrac 45} = \pm \dfrac 2{\sqrt 5}\\\\\\\text{Since}~ \theta ~ \text{lies in the second quadrant,} \cot \theta ~\text{will be negative.}\\\\\text{Hence,}~ \cot \theta = - \dfrac{2}{\sqrt 5}$

6 0

2 years ago

Read 2 more answers

Can someone plz help me ASAP??? my hw in hard and I don't get it

Fynjy0 [20]

9514 1404 393

Answer:

see attachments

Step-by-step explanation:

These are "one-step" equations.

To solve the ones involving addition or subtraction, identify the constant on the same side with the variable. Add its opposite to both sides.

To solve the ones involving multiplication or division, identify the coefficient of the variable. Multiply both sides by its reciprocal (or, equivalently, divide by that coefficient).

Here are more details for the problems on the first page.

w +8 = -3 ⇒ w = -3 -8 = -11
z +(-7) = -20 ⇒ z = -20 +7 = -13
1/5x = 11 ⇒ x = 5·11 = 55
1u = -88 ⇒ u = -88/11 = -8
1/9y = -2 ⇒ y = 9(-2) = -18
12 +n = 7 ⇒ n = 7 -12 = -5
q +(-4) = 8 ⇒ q = 8 +4 = 12

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The second page is solved the same way. It is shown in the second attachment.

7 0

3 years ago