prove this identity:
secθ +1/tanθ = tanθ/secθ −1
To prove the given trigonometric identity:
We'll start with the left-hand side (LHS):
LHS = secθ + 1/tanθ
Step 1: Rewrite everything in terms of sinθ and cosθ:
secθ = 1/cosθ
tanθ = sinθ/cosθ
LHS = 1/cosθ + 1/(sinθ/cosθ)
= 1/cosθ + cosθ/sinθ (by multiplying the second fraction by (1/cosθ))
Step 2: Find a common denominator for the two fractions:
The common denominator will be sinθ * cosθ.
LHS = (sinθ + cos²θ)/(cosθ * sinθ)
= (sinθ + cos²θ)/sinθcosθ
Now, let's work on the right-hand side.
RHS = tanθ/secθ - 1
Step 1: Rewrite tanθ and secθ in terms of sinθ and cosθ:
tanθ = sinθ/cosθ
secθ = 1/cosθ
RHS = (sinθ/cosθ) / (1/cosθ) - 1
= (sinθ/cosθ) * cosθ - 1 (by multiplying the second fraction by (cosθ/cosθ))
= sinθ - 1
Now, we need to show that the LHS equals the RHS.
Therefore, we need to prove that:
(sinθ + cos²θ)/(sinθcosθ) = sinθ - 1
Step 1: Simplify the right side:
sinθ - 1 = (sinθ * cosθ - cosθ)/cosθ
= (sinθ * cosθ - cosθ * cosθ)/cosθ
= (cosθ * (sinθ - cosθ))/cosθ
= sinθ - cosθ (by canceling out cosθ)
Now, it is clear that both sides are equal since LHS = sinθ - cosθ = RHS.
Hence, we have proven the given trigonometric identity.