Question

If θ is an acute angle and tanθ+cotθ=2, find the value of tan²θ+cot²θ
.

Answer 1

Step-by-step explanation:

Note that

$( \tan( \alpha ) + \cot( \alpha ) ) {}^{2} = \tan {}^{2} ( \alpha ) + \cot {}^{2} ( \alpha ) - 2 \tan( \alpha ) \cot( \alpha ) )$

Note that tan and cot are reciprocals meaning that they cancel our to 1 since we are multiplying them.

For example, the fractions, 1/2 and 2/1 are reciprocals so

$\frac{1}{2} \times \frac{2}{1} = 1$

Therefore, we get

$(2) {}^{2} = \tan {}^{2} ( \alpha ) + \cot {}^{2} ( \alpha ) -2$

$6 = \tan {}^{2} ( \alpha ) + \cot {}^{2} ( \alpha )$

Answer 2

Answer:  2

Work Shown

$\tan \theta + \cot \theta = 2\\\\(\tan \theta + \cot \theta)^2 = 2^2\\\\\tan^2 \theta + 2\tan \theta * \cot \theta + \cot^2 \theta = 4\\\\\tan^2 \theta + 2\tan \theta * \frac{1}{\tan \theta} + \cot^2 \theta = 4$

$\tan^2 \theta + 2*1 + \cot^2 \theta = 4\\\\\tan^2 \theta + 2 + \cot^2 \theta = 4\\\\\tan^2 \theta + \cot^2 \theta = 4-2\\\\\tan^2 \theta + \cot^2 \theta = 2$

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Another approach

Let $\text{x} = \tan \theta$

Then $\cot \theta = \frac{1}{\tan \theta} = \frac{1}{\text{x}}$

If you solved the equation $\text{x} + \frac{1}{\text{x}} = 2$ for x, you should get the solution x = 1. I'll skip the steps to showing how to solve for x.

So,

$\tan^2 \theta + \cot^2 \theta = \tan^2 \theta + \frac{1}{\tan^2 \theta}\\\\\tan^2 \theta + \cot^2 \theta = \text{x}^2 + \frac{1}{\text{x}^2}\\\\\tan^2 \theta + \cot^2 \theta = 1^2 + \frac{1}{1^2}\\\\\tan^2 \theta + \cot^2 \theta = 2$

In short

$\tan^2 \theta + \cot^2 \theta = 2$