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Prove that Sin^2(90-θ)(1+cot^2(90-θ)) = 1.

A. Trigonometric Identity
B. Algebraic Expression
C. Geometric Theorem
D. Calculus Formula

User Manki
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2 Answers

4 votes

Final answer:

The expression Sin^2(90-θ)(1+cot^2(90-θ)) = 1 is a trigonometric identity that can be proved by using fundamental trigonometric relationships and simplifying using the Pythagorean identity for tangent.

Step-by-step explanation:

The question asks to prove the trigonometric identity Sin^2(90-θ)(1+cot^2(90-θ)) = 1. To begin, we use the fundamental trigonometric identities that relate the sine and cosine functions. Since sin(90-θ) = cos(θ) and cot(90-θ) = tan(θ), the expression can be rewritten as cos^2(θ)(1+tan^2(θ)). The Pythagorean identity for tangent states that 1+tan^2(θ) = sec^2(θ), which means we can transform the expression into cos^2(θ)sec^2(θ). Since sec(θ) = 1/cos(θ), it simplifies to cos^2(θ)(1/cos^2(θ)), which is equal to 1, as cos^2(θ) in the numerator and denominator cancel out each other.

User Doogie
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1 vote

Final answer:

The trigonometric expression Sin^2(90-θ)(1+cot^2(90-θ)) simplifies to 1 using the co-function and Pythagorean identities which involve transforming the given terms into cosine and tangent functions and demonstrating that their product is equal to 1.

Step-by-step explanation:

To prove that Sin2(90-θ)(1+cot2(90-θ)) = 1, we can use basic trigonometric identities. First, we recognize that sin(90-θ) is the same as cos(θ). This follows from the co-function identity where sin(90°-x) = cos(x). So, the expression can be rewritten as cos2(θ).

Next, cot(90-θ) is the same as tan(θ) and cot2(θ) is the same as tan2(θ), based on the co-function identity again where cot(90°-x) = tan(x). So, we now have cos2(θ) * (1 + tan2(θ)). Now by the Pythagorean identity, 1 + tan2(θ) is equal to sec2(θ).

Now our expression is cos2(θ) * sec2(θ). But by definition, sec(θ) is 1/cos(θ), hence sec2(θ) is 1/cos2(θ). Multiplying cos2(θ) with 1/cos2(θ), we get 1. Therefore, the given expression simplifies to 1, completing the proof.

User Kiryl Ivanou
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