# 3.2.3: Proofs of Trigonometric Identities

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Convert to sine/cosine, use basic identities, and simplify sides of the equation.

Verify that $$\dfrac{\sin^2 x}{\tan^2 x}=1−\sin^2 x$$.

## Verifying Trigonometric Identities

Now that you are comfortable simplifying expressions, we will extend the idea to verifying entire identities. Here are a few helpful hints to verify an identity:

• Change everything into terms of sine and cosine.
• Use the identities when you can.
• Start with simplifying the left-hand side of the equation, then, once you get stuck, simplify the right-hand side. As long as the two sides end up with the same final expression, the identity is true.

Let's verify the following identities.

1. $$\dfrac{\cot^2 x}{\csc x}=\csc x−\sin x$$

Rather than have an equal sign between the two sides of the equation, we will draw a vertical line so that it is easier to see what we do to each side of the equation. Start with changing everything into sine and cosine.

$$\begin{array}{l|l} \dfrac{\cot ^{2} x}{\csc x} & \csc x-\sin x \\ \dfrac{\dfrac{\cos ^{2} x}{\sin ^{2} x}}{\dfrac{1}{\sin x}} & \dfrac{1}{\sin x} \\ \dfrac{\cos ^{2} x}{\sin x} & \end{array}$$

Now, it looks like we are at an impasse with the left-hand side. Let’s combine the right-hand side by giving them same denominator.

$$\begin{array}{|c} \dfrac{1}{\sin x}-\dfrac{\sin ^{2} x}{\sin x} \\ \dfrac{1-\sin ^{2} x}{\sin x} \\ \dfrac{\cos ^{2} x}{\sin x} \end{array}$$

The two sides reduce to the same expression, so we can conclude this is a valid identity. In the last step, we used the Pythagorean Identity, $$\sin^2 \theta +\cos^2 \theta =1$$, and isolated the $$\cos^2 x=1−\sin^2x$$.

There are usually more than one way to verify a trig identity. When proving this identity in the first step, rather than changing the cotangent to$$\dfrac{\cos^2 x}{\sin^2 x}$$, we could have also substituted the identity $$\cot^2 x=\csc^2 x−1$$.

1. $$\dfrac{\sin x}{1−\cos x}=\dfrac{1+\cos x}{\sin x}$$

Multiply the left-hand side of the equation by $$\dfrac{1+\cos x}{1+\cos x}$$.

\begin{aligned} \dfrac{\sin x}{1-\cos x} &=\dfrac{1+\cos x}{\sin x} \\ \dfrac{1+\cos x}{1+\cos x} \cdot \dfrac{\sin x}{1-\cos x} &=\\ \dfrac{\sin (1+\cos x)}{1-\cos ^{2} x} &=\\ \dfrac{\sin (1+\cos x)}{\sin ^{2} x} &=\\ \dfrac{1+\cos x}{\sin x} &= \end{aligned}

The two sides are the same, so we are done.

1. $$\sec(−x)=\sec x$$

Change secant to cosine.

$$\sec(−x)=\dfrac{1}{cos(−x)}$$

From the Negative Angle Identities, we know that $$\cos(−x)=\cos x$$.

\begin{aligned} &=\dfrac{1}{\cos x} \\&=\sec x\end{aligned}

##### Example $$\PageIndex{1}$$

Earlier, you were asked to verify that $$\sin^2 x \tan^2 x=1−\sin^2 x$$.

Solution

Start by simplifying the left-hand side of the equation.

$$\sin ^2 x \tan ^2 x=\dfrac{\sin ^2 x}{\dfrac{\sin ^2 x}{\cos ^2 x}}=\cos ^2 x$$

Now simplify the right-hand side of the equation. By manipulating the Trigonometric Identity,

$$\sin ^2 x+\cos ^2 x=1$$, we get $$\cos ^2 x=1−\sin ^2 x$$.

$$\cos ^2 x=\cos ^2 x$$ and the equation is verified.

Verify the following identities.

##### Example $$\PageIndex{2}$$

$$\cos x \sec x=1$$

Solution

Change secant to cosine.

\begin{aligned} \cos x \sec x&=\cos x \cdot \dfrac{1}{\cos x} \\&=1\end{aligned}

##### Example $$\PageIndex{3}$$

$$2−sec^2 x=1−\tan ^2 x$$

Solution

Use the identity $$1+\tan^2 \theta =\sec^2 \theta$$.

\begin{aligned} 2−\sec^2 x &=2−(1+\tan ^2 x) \\&=2−1−\tan ^2 x \\&=1−\tan ^2 x\end{aligned}

##### Example $$\PageIndex{4}$$

$$\dfrac{\cos(−x)}{1+\sin(−x)}=\sec x+\tan x$$

Solution

Here, start with the Negative Angle Identities and multiply the top and bottom by $$\dfrac{1+\sin x}{1+\sin x}$$ to make the denominator a monomial.

\begin{aligned} \dfrac{\cos (-x)}{1+\sin (-x)} &=\dfrac{\cos x}{1-\sin x} \cdot \dfrac{1+\sin x}{1+\sin x} \\ &=\dfrac{\cos x(1+\sin x)}{1-\sin ^{2} x} \\ &=\dfrac{\cos x(1+\sin x)}{\cos ^{2} x} \\ &=\dfrac{1+\sin x}{\cos x} \\ &=\dfrac{1}{\cos x}+\dfrac{\sin x}{\cos x} \\ &=\sec x+\tan x \end{aligned}

## Review

Verify the following identities.

1. $$\cot(−x)=−\cot x$$
2. $$\csc(−x)=−\csc x$$
3. $$\tan x \csc x \cos x=1$$
4. $$\sin x+\cos x \cot x=\csc x$$
5. $$\csc\left(\dfrac{\pi}{2}−x\right)=\sec x$$
6. $$\tan\left(\dfrac{\pi}{2}−x\right)=\tan x$$
7. $$\dfrac{\csc x}{\sin x}−\dfrac{\cot x}{\tan x}=1$$
8. $$\dfrac{\tan ^2 x}{\tan ^2 x+1}=\sin ^2 x$$
9. $$(\sin x−\cos x)^2+(\sin x+\cos x)^2=2$$
10. $$\sin x−\sin x \cos ^2 x=\sin ^3 x$$
11. $$\tan ^2 x+1+\tan x \sec x=\dfrac{1+\sin x}{\cos ^2 x}$$
12. $$\cos ^2 x=\csc x \cos x \tan x+\cot x$$
13. $$\dfrac{1}{1−\sin x}−\dfrac{1}{1+\sin x}=2\tan x \sec x$$
14. $$\csc^4 x−\cot^4 x=\csc ^2 x+\cot^2x$$
15. $$(\sin x−\tan x)(\cos x−\cot x)=(\sin x−1)(\cos x−1)$$