Trigonometry - Identities of Inverse Operations

Simplify using identities. Leave no fractions in your answer.

$\sec \theta(cos\theta\sin\theta) - \csc\theta\tan\theta$

$\sin\theta - \sec\theta$

$\cot\theta$

$\cos\theta - \sec\theta$

$\sin\theta - \cos\theta$

$\cos\theta-\sin\theta$

Explanation

The easiest first step is to simplify our inverse identities:

$\sec \theta(cos\theta\sin\theta) - \csc\theta\tan\theta = \bigg(\frac{1}{cos\theta}\bigg)cos\theta\sin\theta-\bigg(\frac{1}{\sin\theta}\bigg)\bigg(\frac{sin\theta}{\cos\theta}\bigg)$

Cross cancelling, we end up with

$\bigg(\frac{1}{cos\theta}\bigg)cos\theta\sin\theta-\bigg(\frac{1}{\sin\theta}\bigg)\bigg(\frac{\sin\theta}{\cos\theta}\bigg)=\sin\theta-\frac{1}{\cos\theta}$

Finally, eliminate the fraction:

$\sin\theta - \frac{1}{\cos\theta}= \sin\theta-\sec\theta$

Thus,

$\sec \theta(cos\theta\sin\theta) - \csc\theta\tan\theta = \sin\theta - \sec\theta$

Simplify each expression below. Your answer should have (at most) one trigonometric function and no fractions.

1. $\frac{cot(x)}{tan(x)}$

$\frac{cos^2(x)}{sin^2(x)}$

Explanation

Using the quotient identities for trig functions, you can rewrite,

$cot(x)=\frac{cos(x)}{sin(x)}$

and

$tan(x)=\frac{sin(x)}{cos(x)}$

Then the fraction becomes

$\frac{cot(x)}{tan(x)}=\frac{cos^2(x)}{sin^2(x)}=cot^2(x)$

Simplify each expression below. Your answer should have (at most) one trigonometric function and no fractions.

Explanation

Use the Pythagorean Identities:

and

Thus the expression becomes,

Simplify each expression below. Your answer should have (at most) one trigonometric function and no fractions.

Explanation

Use the distributive property (FOIL method) to simplify the expression.

Using Pythagorean Identities:

Simplify each expression below. Your answer should have (at most) one trigonometric function and no fractions.

$(sin^2(x)+cos^2(x)) - \frac{cos^2(x)}{(1-sin(x))}$

Explanation

First, simplify the first term in the expression to 1 because of the Pythagorean Identity.

Then, simplify the second term to

$\frac{(1 - sin^2x)}{(1 - sinx)}$ .

This reduces to

$\frac{(1-sinx)(1+sinx)}{1-sinx}=1+sin(x)$ .

The expression is now,

Distribute the negative and get,

Solve each question over the interval $0 < x < 2\pi$

$\frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$

$\frac{\pi}{2}, \frac{3\pi}{2}, 2\pi$

$\frac{\pi}{4}, \frac{5\pi}{4}$

$0, 2\pi$

Explanation

Divide both sides by to get .

Take the square root of both sides to get that and .

The angles for which this is true (this is taking the arctan) are every angle when and .

These angles are all the multiples of $\frac{\pi}{4}$ .

can be stated as all of the following except...

$\frac{sin(x)}{cos(x)}$

$cot(\frac{\pi}{2} - x)$

$cot(x)^{-1}$

Explanation

Let's look at these individually:

$\frac{sin(x)}{cos(x)}$ is true by definition, as is $cot(x)^{-1}$ .

$cot(\frac{\pi}{2} - x)$ is also true because of a co-function identity.

This leaves two - and we can tell which of these does not work using the fact that , which means that is our answer.

Identities of Inverse Operations

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