Question

if tanΘ + cotΘ = 2 find tan² + cot²θ

Answer 1

$\red{\bold{Answer:}}$

$given \: that \: \\ tanθ + cotθ = 2 \\ on \: squaring \: \\ {(tanθ + cotθ)}^{2} = {2}^{2} \\ {tan}^{2} θ + 2tanθcotθ + {cot}^{2} θ = 4 \\ {tan}^{2} θ + {cot}^{2} θ + 2 = 4 \\ {tan}^{2} θ + {cot}^{2} θ = 4 - 2 \\{tan}^{2} θ + {cot}^{2} θ = 2$

Your answer is 2

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Answer 2

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Answer:

2

Step-by-step explanation:

$tanx + cotx = 2 \\ {(tanx + cotx)}^{2} = {2}^{2} \\ {tan}^{2} x + {cot}^{2} x + 2tanxcotx = 4 \\ {tan}^{2} x + {cot}^{2} x + 2 \times \frac{sinx}{cosx} \times \frac{cosx}{sinx} = 4 \\ {tan}^{2} x + {cot}^{2} x = 4 - 2 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 2$