The value of cot(4π+θ).cot(4π−θ) is

⟶cot(4π−θ).cot(4π+θ)=tan(4π−θ)1×tan(4π+θ)1=1−tanθ1+tanθ×(1+tanθ1−tanθ)=1

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The value of ({}{4}+).({}{4}-) is

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The value of $cot (\frac{π}{4} + θ) . cot (\frac{π}{4} - θ)$ is

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⟶ cot (\frac{π}{4} - θ) . cot (\frac{π}{4} + θ)

= \frac{1}{tan ( \frac{π}{4} - θ )} \times \frac{1}{tan ( \frac{π}{4} + θ )}

= \frac{1 + tan θ}{1 - tan θ} \times (\frac{1 - tan θ}{1 + tan θ})

= 1

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Question Text	The value of $cot (\frac{π}{4} + θ) . cot (\frac{π}{4} - θ)$ is
Answer Type	Text solution:1
Upvotes	150