# Proving the trigonometric identities $\tan x + \cot x = 2\csc 2x$ and $\sec^2 \frac{x}{2} = \frac{2}{1 + \cos x}$

I need help with identities. I don't really know how to start or if I'm doing it right. Can someone show me the steps in solving these?

1.) $\tan x+ \cot x= 2\csc 2x$

$$\frac{\sin x}{\cos x} + \frac{\cos x}{\sin x}= 2\csc 2x$$

Is this the right way?

2.) $\sec^2 \dfrac{x}{2} = \dfrac{2}{1+ \cos x}$

I'm not sure how to start this one.

• Hint: For the number $2$ remember that $$\color{blue}{\sec t=\frac1{\cos t}\qquad \text{and}\qquad\cos^2\alpha=\frac{1+\cos 2\alpha}2}$$ Then $$\sec^2\frac x2=\frac1{\cos^2\frac x2}= ?$$ – Ángel Mario Gallegos Apr 18 '17 at 6:04
• Please read this tutorial on how to typeset mathematics on this site. For the first one, get a common denominator, then use the trigonometric identity $\sin^2x + \cos^2x = 1$. You will also need the trigonometric identity $\sin 2x = 2\sin x\cos x$. – N. F. Taussig Apr 18 '17 at 7:15

## 2 Answers

I would say that writing $$\frac{\sin x}{\cos x} + \frac{\cos x}{\sin x}= 2\csc 2x$$ is certainly not the way to proceed. You are in essence assuming what you need to prove. I would start proving this by writing $$\tan x+\cot x= \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} =\frac{\sin^2 x+\cos^2x}{\sin x\cos x}=\cdots.$$

For $1.)$ use the definition of tangent, cotangent, cosecant and the double angle formula for sine.

$$\csc 2x=\frac{1}{\sin2x}~~~~~~\textrm{and}~~~~~~~~\sin 2x=2\sin x\cos x$$ hence $$1-2\sin x\cos x\csc 2x=0\tag 1$$ as well as $$\cot x=\frac{\cos x}{\sin x} ~~~~~~\textrm{and}~~~~~~~~\tan x=\frac{\sin x}{\cos x}$$ or $$\cos x=\sin x\cot x \tag 2$$ $$\sin x=\cos x\tan x \tag 3$$ Starting from $(1)$ $$\Leftrightarrow\sin^2 x + \cos^2 x -2\sin x \cos x \csc 2x=0$$ use $(2)$ $$\Leftrightarrow\sin^2 x + \cos\sin x\cot x -2\sin x \cos x \csc 2x=0$$ divide by $\sin x \neq 0$ $$\Rightarrow\sin x + \cos x\cot x -2\cos x \csc 2x=0$$ use (3) $$\Leftrightarrow\cos x\tan x+ \cos x\cot x -2\cos x \csc 2x=0$$ divide by $\cos x \neq 0$ $$\Rightarrow\tan x+ \cot x -2\csc 2x=0\Leftrightarrow\tan x+ \cot x=2\csc 2x$$ For $2.)$ substitute $x=2y$, use the definition of secant and the double angle formula for cosine. Definition of secant $$\sec y\cos y=1\tag 1$$ and the double angle formula for cosine $$\cos 2y-2\cos^2y+1=0$$ multiply by $\sec^2 y\neq 0$ $$\Rightarrow\sec^2y+\sec^2 y\cos 2y-2\sec^2 y\cos^2 y=0$$ using $(1)$ twice $$\Leftrightarrow\sec^2y+\sec^2 y\cos 2y-2=0$$ collect $\sec^2 y$ $$\Leftrightarrow\sec^2y(1+\cos 2y)-2=0$$ or for $1+\cos x\neq=0$ $$\Rightarrow\sec^2\frac{x}{2}=\frac{2}{1+\cos x}$$