Answer:
-6x-16
Step-by-step explanation:
Answer: 63
Step-by-step explanation:
PEMDAS
<span> I am assuming you want to prove:
csc(x)/[1 - cos(x)] = [1 + cos(x)]/sin^3(x).
</span>
<span>If we multiply the LHS by [1 + cos(x)]/[1 + cos(x)], we get:
LHS = csc(x)/[1 - cos(x)]
= {csc(x)[1 + cos(x)]/{[1 + cos(x)][1 - cos(x)]}
= {csc(x)[1 + cos(x)]}/[1 - cos^2(x)], via difference of squares
= {csc(x)[1 + cos(x)]}/sin^2(x), since sin^2(x) = 1 - cos^2(x).
</span>
<span>Then, since csc(x) = 1/sin(x):
LHS = {csc(x)[1 + cos(x)]}/sin^2(x)
= {[1 + cos(x)]/sin(x)}/sin^2(x)
= [1 + cos(x)]/sin^3(x)
= RHS.
</span>
<span>I hope this helps! </span>
Answer:
need points
Step-by-step explanation:
fjtgykghkchnckh
Yo sup??
The two triangles are similar by AAA property ie
Angle Angle Angle property
Let us name the two triangles as ABC and DEF
By observing the figure
mB=mE=90 (right angled triangle)
mA=mD (parallel lines property)
mC=mF (parallel lines property)
Therefore by AAA the two triangles are similar .
The diagram is given below.
PS: All the credits for image goes to Picasso-PSN03
