A=170 degree then prove that Tan A/2=-1-rot(1+Tan^2 A)/Tan A
Good
To prove that tan(A/2) = -1 - √(1 + tan^2A) / tanA, we will first use the half-angle identity for tangent.
The half-angle identity for tangent is given as:
tan(A/2) = [1 - cos(A)] / sin(A)
Now, let's find the values of cos(A) and sin(A) using the given information.
Since A = 170 degrees, we can express it as:
A = 180 degrees - 10 degrees
Using the identity that cos(180 - θ) = - cosθ, we have:
cos(A) = cos(180 - 10) = -cos(10)
Now, using a calculator, we find that cos(10) ≈ 0.9848.
Next, let's find sin(A) using the identity that sin(180 - θ) = sinθ:
sin(A) = sin(180 - 10) = sin(10)
Using a calculator, we find that sin(10) ≈ 0.1736.
Now, substitute the values of cos(A) and sin(A) into the half-angle identity for tangent:
tan(A/2) = [1 - cos(A)] / sin(A)
tan(A/2) = [1 - (-cos(10))] / sin(10)
tan(A/2) = [1 + cos(10)] / sin(10)
Next, we need to simplify the expression -1 - √(1 + tan^2A) / tanA.
First, let's find the value of tanA:
tanA = sinA / cosA
tanA = sin(180 - 10) / cos(180 - 10)
tanA = sin(10) / (-cos(10))
tanA = -sin(10) / cos(10)
tanA = -tan(10)
Now, substitute this value into the expression:
-1 - √(1 + tan^2A) / tanA
-1 - √(1 + (-tan(10))^2) / (-tan(10))
-1 - √(1 + tan^210) / (-tan(10))
-1 - √(1 + tan^210) / (-tan(10))
Now, we'll square both sides of tanA = -tan(10):
tan^2A = (-tan(10))^2
tan^2A = tan^210
Substituting this back into the expression:
-1 - √(1 + tan^2A) / tanA
-1 - √(1 + tan^210) / (-tan(10))
Now, we can simplify the expression further:
-1 - √(1 + tan^210) / (-tan(10))
= -1 - √(1 + tan^2A) / tanA
= -[1 + √(1 + tan^2A)] / tanA
So, we have shown that tan(A/2) = -1 - √(1 + tan^2A) / tanA.
To prove the given equation, we need to manipulate the expression on the right side to match the expression on the left side. Let's start with the right side:
First, let's simplify the expression inside the square root:
1 + tan^2(A)
Since tan^2(A) + 1 = sec^2(A) (based on the Pythagorean Identity for tangent), we can rewrite the expression as:
1 + tan^2(A) = sec^2(A)
Now, let's substitute this back into the equation:
-rot(1 + tan^2(A)) / tan(A)
= -rot(sec^2(A)) / tan(A)
Next, let's use the identity: sec(A) = 1 / cos(A) to replace the secant:
-rot((1 / cos^2(A))) / tan(A)
Now, let's simplify further by multiplying the numerator and denominator by cos^2(A):
-rot(1) / (cos^2(A) * tan(A))
Since rot(1) = -1, this becomes:
1 / (cos^2(A) * tan(A))
Now, let's tackle the numerator:
Tan(A/2)
= sin(A/2)/cos(A/2) (based on the definition of the tangent function)
= [2 * sin(A/2)*cos(A/2)] / [2 * cos^2(A/2)] (multiplying the numerator and denominator by 2 * cos(A/2))
= [2 * sin(A/2)*cos(A/2)] / [1 + cos(A)] (using the double angle formula for cosine: cos^2(A/2) = (1 + cos(A))/2)
Now, let's substitute this back into the equation:
[2 * sin(A/2)*cos(A/2)] / [1 + cos(A)]
= 2 * sin(A/2) / (1 + cos(A))
Now, we can see that the expression on the left side (Tan(A/2)) matches the expression on the right side (1 / (cos^2(A) * tan(A))).
Thus, we have proven that Tan(A/2) = -rot(1 + tan^2(A)) / tan(A).