Question

$if \: \tan\theta + \cot \theta = 2 \\ then \: \: { \tan}^{2} \theta + { \cot }^{2} \theta = \: ? \: \\ \\ a) \: 0 \: \\ b) \: 1 \\ c) \: 2 \: \\ d) \: 3$

Answer 1

We are given :-

$\tan( \theta) \: + \: \cot( \theta) \: = \: 2$


We know that :-

$= > \tan ^{2} ( \theta) \: + \: \cot^{2} (\theta) \: = (\: \tan( \theta) \: + \: \cot( \theta) ) {}^{2} \: - \: 2 \tan( \theta) \cot( \theta) \\ = 2 ^{2} \: - \: 2 \: = \: 2$

Hope this helps.....

Answer 2

Given : tan Ф + cot Ф = 2

⇒  tan Ф + (1/tan Ф) = 2

⇒ tan^2 Ф + 1 = 2 tan Ф

⇒ tan^2 Ф - 2 tan Ф + 1 = 0

⇒ tan^2 Ф - 2(tanФ)(1) + (1)^2 = 0

We know that a^2 - 2ab + b^2 = (a - b)^2

⇒ (tan Ф - 1)^2 = 0

⇒ tan Ф = 1.

Now,

⇒ tan^2 Ф + cot^2 Ф

⇒ (1) + (1/1)

⇒ 1 + 1

⇒ 2.

Therefore, the answer is 2 - Option (C).

Hope this helps!