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Prove that sinθ cosθ = cotθ is not a trigonometric identity by producing a counterexample.

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Answer:

cmon man

Explanation:

We want to show that sinθ cosθ = cotθ is not true for all values of θ. To do this, we just need to find one counterexample, i.e., one value of θ for which the equation is not true.

Recall that cotθ = cosθ/sinθ. So, the equation sinθ cosθ = cotθ can be rewritten as sinθ cosθ = cosθ/sinθ.

Multiplying both sides by sinθ, we get:

sin^2θ cosθ = cosθ

Dividing both sides by cosθ, we get:

sin^2θ = 1

Taking the square root of both sides, we get:

sinθ = ±1

So, we need to find a value of θ such that sinθ is equal to ±1. This occurs when θ = π/2 + kπ, where k is an integer.

Now, let's check whether sinθ cosθ = cotθ is true for this value of θ. We have:

sin(π/2 + kπ) = ±1

cos(π/2 + kπ) = 0

cot(π/2 + kπ) = undefined (since cos(π/2 + kπ) = 0)

Therefore, sinθ cosθ = cotθ is not true for θ = π/2 + kπ, and we have found a counterexample. This shows that sinθ cosθ = cotθ is not a trigonometric identity.

User Vitomadio
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