How do I prove the identities of these questions?

Question

How do I prove the identities of these questions?

$$ \frac{1+\tan^2 u}{1-\tan^2u}= \frac{1}{\cos^2u-\sin^2u} \tag1 \\ $$ $$ \frac{\sin x + \sin(5x)}{\cos x+\cos(5x)} = \tan(3x) \tag2 $$

Do I substitute the value for the sin or tan first when solving this? I'm confused.

Answer 1

You will need two facts, first the definition of tangent, second main trigonometric identity: $$ \tan x= \frac{\sin x}{\cos x} \\ \sin^2x + \cos^2x = 1 $$

Using that you may transform $(1)$ as: $$ \frac{1+\tan^2u}{1-\tan^2u} = \frac{1+\frac{\sin^2u}{\cos^2u}}{1-\frac{\sin^2u}{\cos^2u}} =\\ = \frac{\frac{\cos^2u + \sin^2u}{\cos^2u}}{\frac{\cos^2u - \sin^2u}{\cos^2u}} = \frac{1}{\cos^2u} \cdot \frac{\cos^2u}{\cos^2u-\sin^2u} = \frac{1}{\cos^2u-\sin^2u} $$

The original $(2)$ which is: $$ \sin x + \frac{\sin 5x}{\cos x + \cos 5x} = \tan 3x $$ is false as shown in another answer (take $x={\pi \over 4}$ for instance). Here is a visualization of LHS and RHS

Answer 2

A very common strategy is to rewrite the expression and take the common denominator. Usually, this leads to an expression which can be simplified more easily.

The most commonly used identity is the Pythagorean Identity, or

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

which is used in almost all such questions. It is also important to note the obvious $\tan x = \frac{\sin x}{\cos x}$, which also helps greatly.

$$\frac{1+\tan^2 u}{1-\tan^2u} = \frac{1}{\cos^2u-\sin^2u} \tag1 \\$$

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

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

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

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

The second identity is just incorrect and false.

$$\color{red}{\sin x+\frac{\sin(5x)}{\cos x+\cos(5x)} \neq \tan(3x)} \tag2$$

Take $x = \frac{\pi}{4}$. Clearly, the RHS becomes $\tan\frac{3\pi}{4}$, which becomes $-1$. However, the LHS is undefined as $\cos\frac{\pi}{4}+\cos\frac{5\pi}{4}$ becomes $0$.

Edit: It has been mentioned in the comment below that perhaps it was a typo on your part. In this case, the following identity has to be proven:

$$\frac{\sin x+\sin(5x)}{\cos x+\cos(5x)} = \tan(3x) \tag{2}$$

Here, you use what are known as the Sum-to-Product identities:

$$\color{blue}{\sin\alpha+\sin\beta = 2\sin\bigg(\frac{\alpha+\beta}{2}\bigg)\cos\bigg(\frac{\alpha-\beta}{2}\bigg)}$$

$$\color{blue}{\cos\alpha+\cos\beta = 2\cos\bigg(\frac{\alpha+\beta}{2}\bigg)\cos\bigg(\frac{\alpha-\beta}{2}\bigg)}$$

Using these, you get

$$\frac{2\sin(3x)\cos(-4x)}{2\cos(3x)\cos(-4x)} = \tan(3x)$$

$$\frac{\sin(3x)}{\cos(3x)} = \tan(3x)$$

$$\tan(3x) = \tan(3x)$$

Answer 3

For the first one: $$\frac{1}{\cos^2(u)-\sin^2(u)} =\frac{\cos^2(u)+\sin^2(u)}{\cos^2(u)-\sin^2(u)}.$$ Now divide numerator and denominator by $\cos^2(u)$.