Proving of this trigonometric identity

Question

$$\frac{\cot \beta}{\csc \beta - 1} + \frac{\cot \beta}{\csc \beta + 1} = 2 \sec \beta$$

What I've done:

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

Right side:

$$2 \sec\beta=\frac{1}{2\cos\beta}$$

Could you tell me where I went wrong? I've tried using a proof program online (symbolab) though those steps are a bit hard for me to follow.

Note: I only want to use what I have on the left side to solve the left side.

Thank you very much.

(Might be some errors first time using math jax..)

Answer 1

You first mistake was listening to your teacher. This has turned you into a $\sin$ / $\cos$ robot, causing you go against your better intuition and find a common denominator. So let's do that now $$\begin{array}{lll} \frac{\cot\beta}{\csc\beta-1}+\frac{\cot\beta}{\csc\beta+1}&=&\frac{\cot\beta}{\csc\beta-1}\cdot\frac{\csc\beta+1}{\csc\beta+1}+\frac{\cot\beta}{\csc\beta+1}\cdot\frac{\csc\beta-1}{\csc\beta-1}\\ &=&\cot\beta\bigg(\frac{\csc\beta+1}{\csc^2\beta-1}+\frac{\csc\beta-1}{\csc^2\beta-1}\bigg)\\ &=&\cot\beta\bigg(\frac{\csc\beta+1+\csc\beta-1}{\csc^2\beta-1}\bigg)\\ &=&\cot\beta\bigg(\frac{2\csc\beta}{\csc^2\beta-1}\bigg)\\ &=&\cot\beta\bigg(\frac{2\csc\beta}{\cot^2\beta}\bigg)\\ &=&2\frac{\csc\beta}{\cot\beta}\\ &=&2\csc\beta\tan\beta\\ \end{array}$$

At this point feel free to do the $\sin$/$\cos$ thingy, use the following identity $$\tan\beta=\frac{\sin\beta}{\cos\beta}=\frac{\sin\beta}{\cos\beta}\cdot \frac{\frac{1}{\sin \beta \cos \beta}}{\frac{1}{\sin \beta \cos \beta}}=\frac{\frac{1}{\cos\beta}}{\frac{1}{\sin\beta}}=\frac{\sec\beta}{\csc\beta}$$ Continuing we have $$2\csc\beta\tan\beta=2\csc\beta\cdot\frac{\sec\beta}{\csc\beta}=2\sec\beta$$ without the mess.

Answer 2

The mistake is in the third line of your derivation: the denominators should be $(1-\sin\beta)(1+\sin\beta)$, or $1-\sin^2\beta$, instead of $\sin^2\beta$.