5. Prove: Tan beta sin beta + cos beta = sec beta
6. Prove: ( tan y cos^2 y + sin^2 y)/ (sin y) = cos y+sin y
7. Prove (1+tan theta)/(1-tan theta) = (sec^2+2tan theta) /(1-tan^2 theta)
try using the identity
sec^2 x = 1 + tan^2 x
It should get you through most of these
To prove these trigonometric identities, we can use various trigonometric identities and algebraic manipulations. Let's start with the first identity:
5. Proving: tan(beta) * sin(beta) + cos(beta) = sec(beta)
We'll use the definitions of tangent, sine, cosine, and secant:
tan(beta) = sin(beta) / cos(beta)
sec(beta) = 1 / cos(beta)
Starting with the left-hand side (LHS):
LHS = tan(beta) * sin(beta) + cos(beta)
= (sin(beta) / cos(beta)) * sin(beta) + cos(beta)
= sin^2(beta) / cos(beta) + cos(beta)
To combine the terms on the LHS, we'll find a common denominator:
LHS = (sin^2(beta) + cos^2(beta)) / cos(beta)
= 1 / cos(beta) [using the Pythagorean identity: sin^2(beta) + cos^2(beta) = 1]
= sec(beta) [definition of secant]
Therefore, LHS = RHS, and the identity is proven.
Moving on to the next identity:
6. Proving: (tan(y) * cos^2(y) + sin^2(y)) / sin(y) = cos(y) + sin(y)
Starting with the left-hand side (LHS):
LHS = (tan(y) * cos^2(y) + sin^2(y)) / sin(y)
Using the definition of tangent (tan):
LHS = (sin(y) / cos(y)) * cos^2(y) + sin^2(y)) / sin(y)
Expanding and simplifying:
LHS = sin(y) * cos(y) + sin^2(y) / sin(y)
= sin(y) * cos(y) + sin(y)
Factoring out sin(y) from the first term:
LHS = sin(y) * (cos(y) + 1)
Using the Pythagorean identity: cos^2(y) = 1 - sin^2(y), we can simplify the first term as:
cos(y) + 1 = 1 - sin^2(y) + 1
= 2 - sin^2(y)
Substituting it back into LHS:
LHS = sin(y) * (2 - sin^2(y))
= 2sin(y) - sin^3(y)
Now, we simplify the right-hand side (RHS):
RHS = cos(y) + sin(y)
Using the Pythagorean identity: cos^2(y) = 1 - sin^2(y) and substituting it into RHS:
RHS = √(1 - sin^2(y)) + sin(y)
To simplify the RHS, we'll square both sides:
RHS^2 = (1 - sin^2(y)) + 2√(1 - sin^2(y))sin(y) + sin^2(y)
= 1 + 2√(1 - sin^2(y))sin(y)
Taking the square root:
RHS = √(1 + 2√(1 - sin^2(y))sin(y))
We can see that LHS and RHS are not equivalent. Therefore, the given identity is not proven.
Lastly, let's prove the third identity:
7. Proving: (1 + tan(theta)) / (1 - tan(theta)) = (sec^2(theta) + 2tan(theta)) / (1 - tan^2(theta))
Starting with the LHS:
LHS = (1 + tan(theta)) / (1 - tan(theta))
Using the definition of tangent:
LHS = (1 + sin(theta) / cos(theta)) / (1 - sin(theta) / cos(theta))
Finding a common denominator:
LHS = (cos(theta) + sin(theta)) / (cos(theta) - sin(theta))
Next, we'll multiply the numerator and denominator by the conjugate of the denominator, which is (cos(theta) + sin(theta)):
LHS = (cos(theta) + sin(theta)) * (cos(theta) + sin(theta)) / (cos(theta) - sin(theta)) * (cos(theta) + sin(theta))
Expanding and simplifying:
LHS = (cos^2(theta) + 2sin(theta)cos(theta) + sin^2(theta)) / (cos^2(theta) - sin^2(theta))
= (cos^2(theta) - sin^2(theta) + 2sin(theta)cos(theta) + sin^2(theta)) / (cos^2(theta) - sin^2(theta))
= (cos^2(theta) + sin^2(theta) + 2sin(theta)cos(theta)) / (cos^2(theta) - sin^2(theta))
= (1 + sin(theta)cos(theta) / cos^2(theta) - sin^2(theta))
Using the Pythagorean identity: cos^2(theta) = 1 - sin^2(theta), we simplify further:
LHS = (1 + sin(theta)cos(theta) / (1 - sin^2(theta))
Now, we'll work on the RHS:
RHS = (sec^2(theta) + 2tan(theta)) / (1 - tan^2(theta))
Using the definitions of secant and tangent:
RHS = (1 / cos^2(theta) + 2sin(theta) / cos(theta)) / (1 - sin^2(theta) / cos^2(theta))
= (1 + 2sin(theta)cos(theta) / cos^2(theta)) / (cos^2(theta) - sin^2(theta) / cos^2(theta))
= (1 + sin(theta)cos(theta) / cos^2(theta)) / (1 - sin^2(theta))
We can see that the LHS and RHS are identical, so the given identity is proven.