Originally Posted by

**Gusbob** 1. LHS = $\displaystyle (1 + cos \theta)(1-cos \theta)$

This is the difference of two squares, so

= $\displaystyle (1-cos^2 \theta)$

Then use the trigonometric identity $\displaystyle cos^2\theta + sin^2\theta = 1$

= $\displaystyle 1- (1-sin^2\theta)

= sin^2\theta

= RHS$

2.

$\displaystyle tan\theta$ is also $\displaystyle \frac{sin\theta}{cos\theta} $. Inverse tan, (not arctan) is $\displaystyle cot \theta$.

$\displaystyle cot\theta = \frac{1}{tan\theta}

= \frac{cos\theta}{sin\theta}$

Using this you should be able to solve 2.

3.for this question, I'm pretty sure you mean:

$\displaystyle \frac{tan\theta + cot\theta}{tan\theta}$

= $\displaystyle 1 + (cot\theta)( \frac{1}{tan\theta})$

From question two, I've explained that $\displaystyle \frac{1}{tan\theta} = cot\theta$

In this case, use the identity $\displaystyle 1 + cot^2\theta = cosec^2\theta$