show that csc2 theta - tan2 (90 deg - theta) = sin2 theta + sin2 (90 deg - theta)
To prove this trigonometric identity, we'll start by simplifying the left-hand side (LHS) and the right-hand side (RHS) of the equation separately.
Let's simplify the LHS first:
LHS: csc^2θ - tan^2(90° - θ)
Using the trigonometric identity csc^2θ = 1/sin^2θ, we can rewrite csc^2θ as:
LHS: 1/sin^2θ - tan^2(90° - θ)
Now, let's evaluate tan^2(90° - θ). By using the complementary angle identity tan(90° - θ) = cotθ, we can rewrite the expression as:
LHS: 1/sin^2θ - cot^2θ
Using the identity cot^2θ = 1/tan^2θ = (cos^2θ/sin^2θ), we can simplify further:
LHS: 1/sin^2θ - 1/tan^2θ = 1/sin^2θ - cos^2θ/sin^2θ
Next, we need to find a common denominator for the two fractions on the LHS:
LHS: (1 - cos^2θ)/sin^2θ
Using the identity sin^2θ + cos^2θ = 1, we can rewrite (1 - cos^2θ) as sin^2θ:
LHS: sin^2θ/sin^2θ = 1
Now, we'll simplify the RHS of the equation:
RHS: sin^2θ + sin^2(90° - θ)
Using the identity sin^2(90° - θ) = cos^2θ, we can rewrite the expression as:
RHS: sin^2θ + cos^2θ
By the Pythagorean identity sin^2θ + cos^2θ = 1, we find that the RHS is equal to 1.
Since both the LHS and RHS of the equation simplify to 1, we have shown that csc^2θ - tan^2(90° - θ) = sin^2θ + sin^2(90° - θ).