Prove the following identities:
a) (cosec θ + cotθ)(cosecθ - cotθ) = cotθtanθ
b) 2/(1 + sinθ) + 1/(1- sinθ) = (3sec^2)θ - tanθcosecθ
I basically just have to prove Left HandSide = Right Hand Side
(a) You should immediately recognize that the right side equals 1. If you don't, then review the relationship of tan and cot.
Multiply out the two factors on the left and you get
csc^2 θ - cot^2 θ) = 1
Rewrite the left side as
1/ sin^2 - cos^2/sin^2 = 1
which is the same as
(1 - cos^2)/sin^2 = 1
1 = 1
For (b), I suggest first rewriting the left side with a common denominator. Also recognize that tan*csc on the right equals sec = 1/cos
Thanks drwls that really helped !
:)
Sure! Let's prove each identity step by step:
a) (cosec θ + cotθ)(cosecθ - cotθ) = cotθtanθ
To prove this identity, we'll start by expanding the left-hand side (LHS) using the distributive property:
LHS = cosec θ * cosec θ - cosec θ * cot θ + cot θ * cosec θ - cot θ * cot θ
Now, let's simplify each term:
cosec θ * cosec θ = (1/sin θ) * (1/sin θ) = 1/(sin θ * sin θ) = 1/sin² θ = csc² θ
cosec θ * cot θ = (1/sin θ) * (cos θ/sin θ) = cos θ/sin² θ = cot θ/sin θ = cot θcsc θ
cot θ * cosec θ = (cos θ/sin θ) * (1/sin θ) = cos θ/(sin θ * sin θ) = cos θ/sin² θ = cot θ/sin θ = cot θcsc θ
cot θ * cot θ = (cos θ/sin θ) * (cos θ/sin θ) = cos² θ/(sin θ * sin θ) = cos² θ/sin² θ = cot² θ
Now, substitute these simplifications back into the expanded expression:
LHS = csc² θ - cot θcsc θ + cot θcsc θ - cot² θ
Notice that the middle terms will cancel each other:
LHS = csc² θ - cot² θ
Using the Pythagorean Identity csc² θ = 1 + cot² θ:
LHS = 1 + cot² θ - cot² θ
Simplifying further:
LHS = 1
Since the left-hand side simplifies to 1, and the right-hand side is cot θtan θ, which is also equal to 1, we have proven the identity.
b) 2/(1 + sinθ) + 1/(1- sinθ) = (3sec² θ) - tanθcosecθ
To prove this identity, we'll start by manipulating the left-hand side (LHS):
LHS = 2/(1 + sinθ) + 1/(1 - sinθ)
To combine the fractions, find a common denominator, which in this case is (1 + sinθ)(1 - sinθ):
LHS = [2(1 - sinθ) + (1 + sinθ)] / [(1 + sinθ)(1 - sinθ)]
Simplify the numerator:
LHS = [2 - 2sinθ + 1 + sinθ] / [(1 + sinθ)(1 - sinθ)]
Combine like terms:
LHS = [3 - sinθ] / [(1 + sinθ)(1 - sinθ)]
To simplify the denominator further, use the identity (a - b)(a + b) = a² - b²:
LHS = [3 - sinθ] / [1 - sin²θ]
Using the Pythagorean Identity 1 - sin²θ = cos²θ:
LHS = [3 - sinθ] / cos²θ
Rearrange the numerator and denominator:
LHS = (3 - sinθ) / cos²θ
To combine the fractions, we'll need a common denominator for the denominator (right-hand side, RHS). Let's find the common denominator, which is cos²θ:
RHS = (3sec²θ - tanθcosecθ)(cos²θ/cos²θ)
Simplify the RHS:
RHS = (3sec²θ * cos²θ - tanθ * cosecθ * cos²θ) / cos²θ
Using the identity secθ = 1/cosθ and cosecθ = 1/sinθ:
RHS = (3(1/cos²θ) * cos²θ - tanθ * (1/sinθ) * cos²θ) / cos²θ
Simplifying further:
RHS = (3 * 1 - tanθ * cosθ) / cos²θ
RHS = (3 - tanθ * cosθ) / cos²θ
Since the LHS and RHS are now in the same form, we can see that they are equal:
LHS = (3 - sinθ) / cos²θ = (3 - tanθ * cosθ) / cos²θ = RHS
Therefore, we have proven the identity.