# Proving of this trigonometric identity

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

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..)

• The inconsistent use of parenthesis and sign "/" makes your argument rather difficult to follow. – user228113 Dec 4 '15 at 2:46
• @G.Sassatelli how's that? – Nitrate Dec 4 '15 at 3:13
• Definitely better. I took the liberty to improve it. It seems you made a couple mistakes between the second and the third line. For instance, it should have been: $$\frac{\frac{\cos \beta}{\sin\beta}}{\frac{1+\sin\beta}{\sin\beta}}=\frac{\cos\beta}{\sin\beta} \cdot \frac{\sin \beta}{1+\sin \beta}=\cdots$$ – user228113 Dec 4 '15 at 3:23
• You should not, because, indeed $$\frac{\cos \beta}{1-\sin \beta}+\frac{\cos \beta}{1+\sin \beta}=\frac{2}{\cos \beta}$$ which leads us to your other mistake of algebra: $$2\sec \beta=2\left(\sec \beta\right)=2\cdot \left(\frac{1}{\cos \beta}\right)=\frac{2}{\cos \beta}$$ – user228113 Dec 4 '15 at 3:39
• @G.Sassatelli I end up with the same end result as before. $\frac{2cos\beta}{1-cos^2\beta}$ – Nitrate Dec 4 '15 at 3:39

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.
• What's going on with $tan\beta$ how did $sin\beta$ become $\frac{1}{cos\beta}$ and $cos\beta$ too? – Nitrate Dec 4 '15 at 17:01
• $\sin\beta$ didn't become $\frac{1}{\cos\beta}$. It became $\frac{1}{\frac{1}{\sin\beta}}$, and $\frac{1}{\cos\beta}$ became $\frac{\frac{1}{\cos\beta}}{1}$. – John Joy Dec 4 '15 at 17:56
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 $\sin^2\beta$.