# How do you prove (1-cot^2theta)/(1+cot^2theta) + 2cos^2theta = 1?

Jun 7, 2016

#### Explanation:

$\frac{1 - {\cot}^{2} \theta}{1 + {\cot}^{2} \theta} + 2 {\cos}^{2} \theta$

= $\frac{1 - {\cos}^{2} \frac{\theta}{\sin} ^ 2 \theta}{1 + {\cos}^{2} \frac{\theta}{\sin} ^ 2 \theta} + 2 {\cos}^{2} \theta$

= $\frac{\frac{{\sin}^{2} \theta - {\cos}^{2} \theta}{\sin} ^ 2 \theta}{\frac{{\sin}^{2} \theta + {\cos}^{2} \theta}{\sin} ^ 2 \theta} + 2 {\cos}^{2} \theta$

= $\left(\frac{{\sin}^{2} \theta - {\cos}^{2} \theta}{\sin} ^ 2 \theta\right) \times \left({\sin}^{2} \frac{\theta}{{\sin}^{2} \theta + {\cos}^{2} \theta}\right) + 2 {\cos}^{2} \theta$

= $\frac{{\sin}^{2} \theta - {\cos}^{2} \theta}{\cancel{{\sin}^{2} \theta}} \times \frac{\cancel{{\sin}^{2} \theta}}{1} + 2 {\cos}^{2} \theta$

= ${\sin}^{2} \theta - {\cos}^{2} \theta + 2 {\cos}^{2} \theta$

= ${\sin}^{2} \theta + {\cos}^{2} \theta = 1$

Jun 7, 2016

Use trigonometric identities. Proof-writing is a skill, not content-based question.

#### Explanation:

LHS
$= \frac{1 - {\cot}^{2} \theta}{1 + {\cot}^{2} \theta} + 2 {\cos}^{2} \theta$
$= \frac{1 - {\cot}^{2} \theta}{{\csc}^{2} \theta} + 2 {\cos}^{2} \theta$
$= \frac{1}{{\csc}^{2} \theta} - \frac{{\cot}^{2} \theta}{{\csc}^{2} \theta} + 2 {\cos}^{2} \theta$
$= {\sin}^{2} \theta - \frac{{\cos}^{2} \theta}{{\sin}^{2} \theta} \cdot {\sin}^{2} \theta + 2 {\cos}^{2} \theta$
$= {\sin}^{2} \theta - {\cos}^{2} \theta + 2 {\cos}^{2} \theta$
$= {\sin}^{2} \theta + {\cos}^{2} \theta$
$= 1$
= RHS