Prove (cosecθ-cotθ)^2=1-cosθ÷1+cosθ
I will assume that you meant:
(cosecθ-cotθ)^2=(1-cosθ)÷(1+cosθ)
or else the identity is not true.
LS = (1/sinØ - cosØ/sinØ)2
= (1-cosØ)^2 / sin^2 Ø
= (1 - cosØ)(1-cosØ)/( (1 - cosØ)(1 + cosØ) )
= (1 - cosØ)/(1+cosØ)
= RS
To prove the trigonometric identity (cosecθ - cotθ)^2 = (1 - cosθ) ÷ (1 + cosθ), we will start with the left-hand side (LHS) and manipulate it to match the right-hand side (RHS).
LHS: (cosecθ - cotθ)^2
First, let's rewrite cosecθ and cotθ in terms of sinθ and cosθ:
cosecθ = 1 / sinθ
cotθ = cosθ / sinθ
Now substitute these values in the LHS expression:
LHS: (1 / sinθ - cosθ / sinθ)^2
Next, find a common denominator for the terms inside the parentheses:
LHS: [(1 - cosθ) / sinθ]^2
To square the expression, we multiply the numerator and denominator by itself:
LHS: [(1 - cosθ)^2 / sinθ^2]
Now, let's simplify the denominator:
LHS: (1 - cosθ)^2 / (1/sinθ) * (1/sinθ)
Multiplying the denominators together, we get:
LHS: (1 - cosθ)^2 / (1/sinθ)^2
Simplifying further by squaring the denominator:
LHS: (1 - cosθ)^2 / (1/sinθ^2)
Since 1/sinθ^2 is equal to cosec^2θ, we can rewrite this expression as:
LHS: (1 - cosθ)^2 / cosec^2θ
Recall that cosecθ is equal to 1/sinθ, so we can substitute this back in:
LHS: (1 - cosθ)^2 / (1/sinθ)^2
Expanding the numerator by squaring the expression:
LHS: (1 - 2cosθ + cos^2θ) / (1/sinθ)^2
Rearranging the terms inside the parentheses:
LHS: (cos^2θ - 2cosθ + 1) / (1/sinθ)^2
Now, let's simplify the denominator by squaring the reciprocal:
LHS: (cos^2θ - 2cosθ + 1) / (sinθ)^2
We can rewrite cos^2θ as 1 - sin^2θ (using the Pythagorean identity):
LHS: (1 - sin^2θ - 2cosθ + 1) / (sinθ)^2
Combining like terms in the numerator:
LHS: (2 - sin^2θ - 2cosθ) / (sinθ)^2
Now, recall the identity sin^2θ + cos^2θ = 1:
LHS: (2 - (1 - cos^2θ) - 2cosθ) / (sinθ)^2
Simplifying the numerator:
LHS: (2 - 1 + cos^2θ - 2cosθ) / (sinθ)^2
Rearranging the terms in the numerator:
LHS: (cos^2θ - 2cosθ + 1) / (sinθ)^2
Which is equal to the right-hand side (RHS) expression.
Therefore, we have proved that (cosecθ - cotθ)^2 = (1 - cosθ) ÷ (1 + cosθ).