If tan A=a/b prove that 2sec theta + 1/ cos theta+2=Õ a2+b2/b
Please put solution clearly now
To prove that 2sec(theta) + 1/cos(theta) + 2 = Õ(a^2+b^2)/b, we can use the given information that tan(A) = a/b.
First, let's start by expressing sec(theta) and cos(theta) in terms of tan(theta):
sec(theta) = 1/cos(theta)
cos(theta) = 1/sec(theta)
Now, substitute these values into the equation:
2sec(theta) + 1/cos(theta) + 2
= 2(1/cos(theta)) + 1/(1/sec(theta)) + 2
= 2/cos(theta) + sec(theta) + 2
Since sec(theta) = 1/cos(theta), we can substitute sec(theta) in terms of cos(theta):
= 2/cos(theta) + 1/cos(theta) + 2
= (2 + 1)/cos(theta) + 2
= 3/cos(theta) + 2
Now, let's use the given information that tan(A) = a/b to express cos(theta) in terms of a and b:
tan(A) = a/b
=> sin(A)/cos(A) = a/b
=> sin(A) = a/b * cos(A)
Now, using Pythagorean identity, we have:
sin^2(A) + cos^2(A) = 1
Substituting sin(A) = a/b * cos(A):
(a^2/b^2) * cos^2(A) + cos^2(A) = 1
Combining like terms:
[(a^2/b^2) + 1] * cos^2(A) = 1
Now, isolate cos^2(A):
cos^2(A) = 1 / [(a^2/b^2) + 1]
Substitute this value of cos^2(A) into the previous equation:
3/cos(theta) + 2
= 3 / [1 / [(a^2/b^2) + 1]] + 2
Simplifying the expression inside the square brackets:
= 3 * [(a^2/b^2) + 1] + 2
Expanding the expression:
= 3 * (a^2/b^2) + 3 + 2
= (3a^2/b^2) + 5
Finally, substituting Õ(a^2+b^2)/b back into the equation:
(3a^2/b^2) + 5 = Õ(a^2+b^2)/b
Therefore, 2sec(theta) + 1/cos(theta) + 2 = Õ(a^2+b^2)/b is proven.
To prove the equation Õ (a^2 + b^2) / b = 2sec(theta) + 1 / cos(theta) + 2, we need to start with the given information, tan(A) = a/b.
First, let's rewrite the given equation using trigonometric identities:
Õ (a^2 + b^2) / b = 2sec(theta) + 1 / cos(theta) + 2
Next, let's manipulate the right side of the equation using trigonometric identities:
2sec(theta) + 1 / cos(theta) + 2
We know that sec(theta) = 1 / cos(theta), so we can substitute that in:
2(1/cos(theta)) + 1/cos(theta) + 2
Finding a common denominator:
(2 + 1 + 2cos(theta)) / cos(theta)
Combining like terms:
(3 + 2cos(theta)) / cos(theta)
Now, let's relate this to tangent using the Pythagorean identity:
cos^2(theta) + sin^2(theta) = 1
Rearranging the equation:
1 = 1 - sin^2(theta)
cos^2(theta) = 1 - sin^2(theta)
Dividing by cos^2(theta):
1 = (1 - sin^2(theta)) / cos^2(theta)
Using the quotient identity (tan(theta) = sin(theta) / cos(theta)):
1 = (1 - tan^2(theta)) / (1/cos^2(theta))
Simplifying:
1 = (1 - tan^2(theta)) * cos^2(theta)
Expanding the equations:
cos^2(theta) = 1 - tan^2(theta)cos^2(theta)
Rearranging:
tan^2(theta)cos^2(theta) + cos^2(theta) = 1
Dividing by cos^2(theta):
tan^2(theta) + 1 = sec^2(theta)
Now, let's substitute this identity back into our equation:
(3 + 2cos(theta)) / cos(theta) = tan^2(theta) + 1
Using the given information, tan(A) = a/b, we can replace tan^2(theta) with (a/b)^2:
(3 + 2cos(theta)) / cos(theta) = (a/b)^2 + 1
Multiplying both sides by b^2:
b^2(3 + 2cos(theta)) = a^2 + b^2
Expanding:
3b^2 + 2b^2cos(theta) = a^2 + b^2
Rearranging the equation to the desired form:
2b^2cos(theta) + 3b^2 = a^2 + b^2
Õ (a^2 + b^2) / b = 2sec(theta) + 1 / cos(theta) + 2
Therefore, we have proven the given equation.
I will assume that A and θ are the same
Draw a triangle. The legs are a and b, so the hypotenuse is √(a^2+b^2)
Now you know that secθ = 1/cosθ = √(a^2+b^2)/b
I expect you can handle the rest of the algebra...