How do I prove that secθ - tanθsinθ = cosθ
So far I have
LS
1/cosθ - sinθ/cosθ (sinθ)
=1/cosθ -sin^2/cosθ
LS = secθ - tanθsinθ
= 1/cosØ - (sinØ/cosØ)sinØ
= 1/cosØ - sin^2Ø/cosØ
= (1 - sin^2 Ø)/cosØ ---> you had it up to there
= cos^2 Ø/cosØ ---> since sin^2 Ø + cos^2 Ø = 1
= cosØ
= RS
To prove that secθ - tanθsinθ = cosθ, you can start by manipulating the left side of the equation using trigonometric identities. Here are the steps:
Step 1: Start with the left side (LS) of the equation:
LS = secθ - tanθsinθ
Step 2: Use the definitions of secant and tangent:
- Recall that secθ is equal to 1/cosθ
- Recall that tanθ is equal to sinθ/cosθ
Substituting these definitions into the equation, we have:
LS = 1/cosθ - (sinθ/cosθ) * sinθ
Step 3: Simplify the expression:
LS = 1/cosθ - sin^2θ/cosθ
Step 4: Combine the fractions by finding a common denominator:
LS = (1 - sin^2θ)/cosθ
Step 5: Recall the Pythagorean identity sin^2θ + cos^2θ = 1:
- Rearrange the equation: sin^2θ = 1 - cos^2θ
- Substitute this into the LS equation:
LS = (1 - (1 - cos^2θ))/cosθ
Step 6: Simplify the expression further:
LS = (1 - 1 + cos^2θ)/cosθ
LS = cos^2θ/cosθ
LS = cosθ/cosθ
LS = 1
Step 7: Since LS = 1 and RS (right side) = cosθ, we have proved that secθ - tanθsinθ = cosθ.
Therefore, secθ - tanθsinθ is equivalent to cosθ.
To prove the given identity, you need to simplify the left-hand side (LHS) until it is equal to the right-hand side (RHS). Let's continue the simplification process from where you left off:
LHS: 1/cosθ - (sinθ/cosθ)(sinθ)
= 1/cosθ - sin^2θ/cosθ
To combine fractions, you need a common denominator, which in this case is cosθ. Therefore, multiply the first term by cosθ/cosθ:
LHS: (1⋅cosθ)/(cosθ⋅cosθ) - sin^2θ/cosθ
= cosθ/(cosθ⋅cosθ) - sin^2θ/cosθ
Now, we can combine the fractions and simplify further:
LHS: (cosθ - sin^2θ)/cosθ
Apply the Pythagorean identity sin^2θ = 1 - cos^2θ:
LHS: (cosθ - (1 - cos^2θ))/cosθ
= (cosθ - 1 + cos^2θ)/cosθ
Rearrange the terms:
LHS: (cos^2θ + cosθ - 1)/cosθ
Observe that the numerator is a quadratic expression. Factorize it:
LHS: [(cosθ - 1)(cosθ + 1)]/cosθ
Simplify the numerator further:
LHS: (cosθ - 1)(cosθ + 1)/cosθ
Now, notice that (cosθ - 1) is the same as -(-cosθ + 1). Rearrange this factor:
LHS: (-1)(-cosθ + 1)(cosθ + 1)/cosθ
Finally, cancel out the -1 terms:
LHS: (cosθ + 1)(-cosθ + 1)/cosθ
Rearrange the terms in the second factor:
LHS: (cosθ + 1)(1 - cosθ)/cosθ
The two factors in the numerator cancel each other out:
LHS: (1 - cosθ)/cosθ
This is now equivalent to the right-hand side (RHS):
RHS: cosθ
Therefore, LHS = RHS, and we have proven that secθ - tanθsinθ = cosθ.