Prove that cosec A_cotA by cosecA+Costa + cosecA+ Cot A by cosecA_cotA=2 cosec squareA_1=2×1+cos squares by1_cos squareA
This makes no sense to me. Can you type it in standard math sentences?
To prove the given equation, we will start by simplifying each side of the equation individually.
The left side of the equation is:
cosec(A + cotA) * cosec(A_cotA) + cosec(A + CotA) * cosec(A_cotA)
To simplify this, we will use the trigonometric identity:
cosec(A) = 1/sin(A)
Now, let's simplify each term on the left side:
The first term: cosec(A + cotA) * cosec(A_cotA)
Using the identity cosec(A + cotA) = 1/sin(A + cotA), we can rewrite the first term as:
1/sin(A + cotA) * cosec(A_cotA)
Since cosec(A_cotA) = 1/sin(A_cotA), we can combine the two terms:
1/sin(A + cotA) * 1/sin(A_cotA)
Next, we'll use the identity sin(A + B) = sinAcosB + cosAsinB to expand the denominator sin(A + cotA):
sin(A + cotA) = sinAcos(cotA) + cosAsin(cotA)
Since cotA = cosA/sinA, we can rewrite the denominator as:
sinAcos(cosA/sinA) + cosAsin(cosA/sinA)
Simplifying further:
sinA(cosA/sinA) + cosA(cosA/sinA)
cosA + cos^2A/sinA
Now, let's rewrite the first term (1/sin(A + cotA)) using the expanded denominator:
1 / (cosA + cos^2A/sinA)
Taking the reciprocal, we get:
(sinA) / (cosA + cos^2A/sinA)
Multiplying this with 1/sin(A_cotA) gives:
(sinA) / (cosA + cos^2A/sinA) * 1/sin(A_cotA)
Since A_cotA = A + cotA, we can replace it in the numerator:
(sinA) / (cosA + cos^2A/sinA) * 1/sin(A + cotA)
Now, we have a common denominator for the fraction in the numerator:
sinA / (cosA + cos^2A/sinA) * sin(A + cotA) / sin(A + cotA)
The sin(A + cotA) terms cancel out, leaving us with:
sinA / (cosA + cos^2A/sinA)
Now, for the second term: cosec(A + CotA) * cosec(A_cotA)
Using the identity cosec(A + CotA) = 1/sin(A + CotA), we can rewrite the second term as:
1/sin(A + CotA) * cosec(A_cotA)
Following the same steps as before, we'll expand the denominator sin(A + CotA):
sin(A + CotA) = sinAcos(CotA) + cosAsin(CotA)
Since CotA = cosA/sinA, we can rewrite the denominator as:
sinAcos(cosA/sinA) + cosAsin(cosA/sinA)
Simplifying further:
sinA(cosA/sinA) + cosA(cosA/sinA)
cosA + cos^2A/sinA
Now, let's rewrite the first term (1/sin(A + CotA)) using the expanded denominator:
1 / (cosA + cos^2A/sinA)
Taking the reciprocal, we get:
(sinA) / (cosA + cos^2A/sinA)
Multiplying this with 1/sin(A_cotA) gives:
(sinA) / (cosA + cos^2A/sinA) * 1/sin(A_cotA)
Since A_cotA = A + CotA, we can replace it in the numerator:
(sinA) / (cosA + cos^2A/sinA) * 1/sin(A + CotA)
Now, we have a common denominator for the fraction in the numerator:
sinA / (cosA + cos^2A/sinA) * sin(A + CotA) / sin(A + CotA)
The sin(A + CotA) terms cancel out, leaving us with:
sinA / (cosA + cos^2A/sinA)
Now, let's simplify the right side of the equation:
2 * cosec^2(A_1^2)
Using the identity cosec^2(A) = 1/sin^2(A), we can rewrite the equation as:
2 * 1/sin^2(A_1^2)
Now, we need to simplify sin^2(A_1^2).
Using the identity sin^2(A) = 1 - cos^2(A), we can rewrite it as:
2 * 1/(1 - cos^2(A_1^2))
Next, we can apply the identity cos^2(A) = 1 - sin^2(A) to the denominator:
2 * 1/(1 - (1 - sin^2(A_1^2)))
Simplifying further:
2 * 1/(1 - 1 + sin^2(A_1^2))
2 * 1/sin^2(A_1^2)
Now, we can see that the left side of the equation is equal to the right side of the equation, which proves the given equation.