prove that (tan theta +cot theta +1 ) (tan theta + cot theta‐1) = sec theta cot theta
pls explain too hard
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It should have been sec²θ+cot²θ instead of secθ + cosθ. It would be incorrect if it was the latter. Considering that you intended to ask (tanθ + cotθ + 1) (tanθ + cotθ - 1) = sec²θ+cot²θ, here's your answer.
Proving:
=> (tanθ + cotθ + 1) (tanθ + cotθ - 1)
We will use the property (a+b)(a-b) = a²-b². Here, tanθ + cotθ is a and 1 is b.
=> (tanθ + cotθ)² - (1)²
Using the property (a+b)² = a² + b² + 2ab, we get
=> tan²θ + cot²θ + 2tanθcotθ - 1
=> (sec²θ - 1) + (cot²θ) + 2tanθ x 1/tanθ - 1
=> sec²θ -1 + cot²θ + 2 - 1
=> sec²θ + cot²θ
Hence proved
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