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Trigonometry · real student question

Prove the identity 1/cot²A + 1/(1 + tan²A) = 1/(1 − sin²A) − 1/csc²A.

Question

Prove that

1cot⁡2A+11+tan⁡2A=11−sin⁡2A−1csc⁡2A.\frac{1}{\cot^{2}A}+\frac{1}{1+\tan^{2}A}=\frac{1}{1-\sin^{2}A}-\frac{1}{\csc^{2}A}.

Step-by-step solution

  1. Choose the strategy: reduce both sides to sines and cosines separately. Neither side is obviously simpler than the other, so instead of transforming one into the other, simplify each independently and show the two results coincide. Every piece here is either a reciprocal identity or a Pythagorean identity.

  2. Simplify the left-hand side. Since cot⁡A=1/tan⁡A\cot A=1/\tan A, the first term is

    1cot⁡2A=tan⁡2A.\frac{1}{\cot^{2}A}=\tan^{2}A.

    For the second, the Pythagorean identity 1+tan⁡2A=sec⁡2A1+\tan^{2}A=\sec^{2}A gives

    11+tan⁡2A=1sec⁡2A=cos⁡2A.\frac{1}{1+\tan^{2}A}=\frac{1}{\sec^{2}A}=\cos^{2}A.

    So the left-hand side is tan⁡2A+cos⁡2A\tan^{2}A+\cos^{2}A.

  3. Simplify the right-hand side. From sin⁡2A+cos⁡2A=1\sin^{2}A+\cos^{2}A=1 we get 1−sin⁡2A=cos⁡2A1-\sin^{2}A=\cos^{2}A, so

    11−sin⁡2A=1cos⁡2A=sec⁡2A,\frac{1}{1-\sin^{2}A}=\frac{1}{\cos^{2}A}=\sec^{2}A,

    and since csc⁡A=1/sin⁡A\csc A=1/\sin A,

    1csc⁡2A=sin⁡2A.\frac{1}{\csc^{2}A}=\sin^{2}A.

    So the right-hand side is sec⁡2A−sin⁡2A\sec^{2}A-\sin^{2}A.

  4. Bridge the two forms. Apply sec⁡2A=1+tan⁡2A\sec^{2}A=1+\tan^{2}A to the right-hand side:

    sec⁡2A−sin⁡2A=1+tan⁡2A−sin⁡2A=tan⁡2A+(1−sin⁡2A)=tan⁡2A+cos⁡2A.\sec^{2}A-\sin^{2}A=1+\tan^{2}A-\sin^{2}A=\tan^{2}A+\bigl(1-\sin^{2}A\bigr)=\tan^{2}A+\cos^{2}A.

    This matches the left-hand side exactly, so the identity holds.

  5. State where it is valid, and test a value. The identity requires cot⁡A\cot A, tan⁡A\tan A and csc⁡A\csc A to be defined and cot⁡A≠0\cot A\ne0, i.e. AA not a multiple of π/2\pi/2. At A=π/3A=\pi/3: the left side is tan⁡2=3\tan^{2}=3 plus cos⁡2=1/4\cos^{2}=1/4, giving 3.253.25; the right side is sec⁡2=4\sec^{2}=4 minus sin⁡2=3/4\sin^{2}=3/4, giving 3.253.25 ✓.

Answer

Both sides equal tan⁡2A+cos⁡2A, so the identity holds for all A≠kπ/2\text{Both sides equal }\tan^{2}A+\cos^{2}A,\text{ so the identity holds for all }A\ne k\pi/2

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