Answer:
Step-by-step explanation:
------(I)
![LHS =\dfrac{Cos \ A}{1+Sin \ A}+\dfrac{1+Sin \ A}{Cos \ A}\\\\\\ = \dfrac{1-Sin \A}{Cos \ A}+\dfrac{1+Sin \ A}{Cos \ A} \ [from \ equation \ (I)]\\\\\\=\dfrac{1-Sin \ A + 1 - Sin \ A}{Cos \ A}\\\\=\dfrac{2}{Cos \ A}\\\\\\=2*Sec \ A = RHS](https://tex.z-dn.net/?f=LHS%20%3D%5Cdfrac%7BCos%20%5C%20A%7D%7B1%2BSin%20%5C%20A%7D%2B%5Cdfrac%7B1%2BSin%20%5C%20A%7D%7BCos%20%5C%20A%7D%5C%5C%5C%5C%5C%5C%20%3D%20%5Cdfrac%7B1-Sin%20%5CA%7D%7BCos%20%5C%20A%7D%2B%5Cdfrac%7B1%2BSin%20%5C%20A%7D%7BCos%20%5C%20A%7D%20%5C%20%5Bfrom%20%5C%20equation%20%5C%20%28I%29%5D%5C%5C%5C%5C%5C%5C%3D%5Cdfrac%7B1-Sin%20%5C%20A%20%2B%201%20-%20Sin%20%5C%20A%7D%7BCos%20%5C%20A%7D%5C%5C%5C%5C%3D%5Cdfrac%7B2%7D%7BCos%20%5C%20A%7D%5C%5C%5C%5C%5C%5C%3D2%2ASec%20%5C%20A%20%3D%20RHS)
We want to put b alone on one side of the equation.
If we multiply by 1/2, we get 1/4bh = A/2. Not only did we not remove a variable from the left side, we made it even more complicated.
So the answer is A.
We need to see all the given answers
Answer:
A = 20sinθ(6 + 5 cosθ) cm²
Step-by-step explanation:
Drop perpendiculars DE and CF to AB.
Then, we have congruent triangles ADE and BCF, plus the rectangle CDEF.
The formula for the area of the trapezium is
A = ½(a + b)h
DE = 10sinθ
AE = 10cosθ
BF = 10cosθ
EF = CD = 12 cm
AB = AE + EF + BF = 10cosθ + 12 + 10 cosθ = 12 + 20cosθ
A = ½(a + b)h
= ½(12 +12 + 20 cosθ) × 10 sinθ
=(24 + 20 cosθ) × 5 sinθ
= 4(6 + 5cosθ) × 5sinθ
= 20sinθ(6 + 5 cosθ) cm²
Supposse that the distance from the point
to the point
is equal to the distance from
to the point
. Then, by the formula of the distnace we must have

cancel the square root and the
's, and then expand the parenthesis to obtain

then, simplifying we obtain

therfore we must have

this means that the points satisfying the propertie must have first component equal to 5. So we can give a lot of examples of such points:
. The set of this points give us a straight line and the points (3,0) and (7,0) are symmetric with respect to this line.