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


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?

  • 7
    $\begingroup$ 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. $\endgroup$ Commented Dec 30, 2018 at 1:09
  • $\begingroup$ @JohnOmielan but that's exactly how the question on textbook is written? $\endgroup$
    – user569622
    Commented Dec 30, 2018 at 1:11
  • 3
    $\begingroup$ Then your textbook must have a typo, because this identity clearly can not be proven true. $\endgroup$ Commented Dec 30, 2018 at 1:12
  • $\begingroup$ @JohnOmielan does that mean every equation that has a dependent variable on one side and a constant on the other side cannot be proven? $\endgroup$
    – user569622
    Commented Dec 30, 2018 at 1:14
  • $\begingroup$ @NobleMushtak I see, thank you for the answer. That must be the case, the book is so old $\endgroup$
    – user569622
    Commented Dec 30, 2018 at 1:15

1 Answer 1


One way we can prove the identity false is as follows:

$$\begin {align} \dfrac {\tan A} {\tan A} + \dfrac{\cot A}{\cot A} = \dfrac {1}{1-2\cos^2 2A} \\ 2 = \dfrac {1}{1-2\cos^2 2A} \\ 2 (1-2\cos^2 2A) = 1 \\ 2 - 4\cos^2 2A = 1 \\ - 4\cos^2 2A = \dfrac {1}{2} \\ \cos^2 2A = -\dfrac {1}{8} \end {align}$$

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.


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