Answer:
<em>Answer is option d</em><em> </em>
<em>Answer is </em><em>given below with explanations</em><em>. </em>
Step-by-step explanation:
We can prove that the two triangles are similar.
We can prove this using AA criterion of similarity.
In triangle DNC and triangle QSC
Vertically opposite angles are equal.
Then Angle QCS = Angle DCN
Two parallel lines cut by a transversal line make the alternate angles are equal.
Then Angle NDC = Angle CQS
By AA criterion of similarity
TRIANGLE DNC ~ TRIANGLE QSC
<em>HAVE A NICE DAY</em><em>!</em>
<em>THANKS FOR GIVING ME THE OPPORTUNITY</em><em> </em><em>TO ANSWER YOUR QUESTION</em><em>. </em>
Have you learned slope yet? Pick two points from the graph and do x-y on the top and x-y on the bottom. That would give you your slope then you can graph it. For example (-2,-1) (-1,1)
-2-(-1)
-1-1
Then that would give you -1/-2 so technically -1/2<< this is what you graph
It is a little less than 90 degrees, which is a right angle, so it will be more acute
=[(sinx/cosx)/(1+1/cosx)] + [(1+1/cosx)/(sinx/cosx)]
=[(sinx/cosx)/(cosx+1/cosx)]+[(cosx+1/cosx)/(sinx/cosx)]
= [sinx/(cosx+1)] + [(cosx+1)/sinx]
= [sin^2x+(cosx+1)^2] / [sinx (cosx+1)]
= [2+2cosx] / [sinx(cosx+1)]
=[2(cosx+1)] / [sinx (cosx+1)]
= 2/sinx
= 2 cscx
(I think this will be helpful for you. if you can see the picture, it has more detail in it.)
Answer:
there is literally nothing there