# How to prove $\frac{\tan (A)}{\tan (A)}+\frac{\cot (A)}{\cot (A)}=\frac{1}{1-2\cos(A)^2}$

I am unable to prove this trigonometric identity

$$\frac{\tan (A)}{\tan (A)}+\frac{\cot (A)}{\cot (A)}=\frac{1}{1-2\cos^2(A)}$$

I have tried to transform the left-hand side and stuck with this

$$\frac{2\sin(A)\cos(A)}{\sin(A)\cos(A)}$$

And I have tried to transform the right-hand side by changing the $$2\cos^2(A)$$ to $$\frac{2}{\sec^2(A)}$$, and used the trigonometric identity $$1+\tan^2(A)=\sec^2(A)$$ and got this instead

$$\frac{1+\tan^2(A)}{\tan^2(A)-1}$$ which I can transform to $$\frac{\cot(A)+\tan(A)}{\tan(A)-\cot(A)}$$.

I cannot get both sides equal, help please?

• I think there's a typo as the left side is always $1 + 1 = 2$, so it's independent of $A$, while the right side is definitely not a constant. – John Omielan Dec 30 '18 at 1:09
• @JohnOmielan but that's exactly how the question on textbook is written? – user569622 Dec 30 '18 at 1:11
• Then your textbook must have a typo, because this identity clearly can not be proven true. – Noble Mushtak Dec 30 '18 at 1:12
• @JohnOmielan does that mean every equation that has a dependent variable on one side and a constant on the other side cannot be proven? – user569622 Dec 30 '18 at 1:14
• @NobleMushtak I see, thank you for the answer. That must be the case, the book is so old – user569622 Dec 30 '18 at 1:15

\begin {align} \dfrac {\tan A} {\tan A} + \dfrac{\cot A}{\cot A} = \dfrac {1}{1-2\cos^2 2A} \\ 2 = \dfrac {1}{1-2\cos^2 2A} \\ 2 (1-2\cos^2 2A) = 1 \\ 2 - 4\cos^2 2A = 1 \\ - 4\cos^2 2A = \dfrac {1}{2} \\ \cos^2 2A = -\dfrac {1}{8} \end {align}
Since the last line would require us to take the square root of a negative number, $$A$$ does not exist, and the identity is false.