Tan θ + Cot θ = 2, then Tan2 θ + Cot2 θ =A) 4 B) 2 C) 6 D) 1

1.	Tan θ + Cot θ = 2, then Tan2 θ + Cot2 θ =A) 4 B) 2 C) 6 D) 1
Answer» Correct option is: B) 2 We have Tan \(\theta\) + Cot \(\theta\) = 2 = \((tan \, \theta + cot \, \theta)^2 = 2^2 = 4\) = \(tan^2\theta + cot^2\theta + 2 \,tan \, \theta \, cot \, \theta = 4 \) (\(\because\) \((a+b)^2 = a^2+b^2 + 2ab\)) = \(tan^2\theta + cot^2\theta + 2 = 4\) (\(\because\) cot \(\theta\) \(\frac 1{tan \, \theta} =\) tan \(\theta\) cot \(\theta\) =1) = \(tan^2\theta + cot^2\theta = 4-2 =2\) Correct option is: B) 2

Answer»

Correct option is: B) 2

We have Tan \(\theta\) + Cot \(\theta\) = 2

= \((tan \, \theta + cot \, \theta)^2 = 2^2 = 4\)

= \(tan^2\theta + cot^2\theta + 2 \,tan \, \theta \, cot \, \theta = 4 \) (\(\because\) \((a+b)^2 = a^2+b^2 + 2ab\))

= \(tan^2\theta + cot^2\theta + 2 = 4\) (\(\because\) cot \(\theta\) \(\frac 1{tan \, \theta} =\) tan \(\theta\) cot \(\theta\) =1)

= \(tan^2\theta + cot^2\theta = 4-2 =2\)

Correct option is: B) 2

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