Is the relation: tan2⁡2θ-tan2⁡θ1-tan2⁡2θtan2⁡θ=tan⁡3θtan⁡θ true?

Given relation is,tan2⁡2θ-tan2⁡θ1-tan2⁡2θtan2⁡θ=tan⁡3θtan⁡θOn taking left hand side,tan2⁡2θ+tan2⁡θ1-tan2⁡2θtan2⁡θ=(tan⁡2θ+tan⁡θ)(tan⁡2θ-tan⁡θ)(1-tan⁡2θtan⁡θ)(1+tan⁡2θtan⁡θ) (Now, a2-b2=(a-b)(a+b))It is known that,tan⁡(A+B)=tan⁡A+tan⁡B1-tan⁡Atan⁡Btan⁡(A-B)=tan⁡A-tan⁡B1+tan⁡Atan⁡BThus, the expression become,=tan⁡2θ+tan⁡θ1-tan⁡2θtan⁡θ×tan⁡2θ-tan⁡θ1+tan⁡2θtan⁡θ=tan⁡(2θ+θ)tan⁡(2θ-θ)=tan⁡3θtan⁡θSo,tan2⁡2θ-tan2⁡θ1-tan2⁡2θtan2⁡θ=tan⁡3θtan⁡θThus, the given relation is true.Option 1 is correct.

Is the relation: tan2⁡2θ-tan2⁡θ1-tan2⁡2θtan2⁡θ=tan⁡3θtan⁡θ true?

Given relation is,tan2⁡2θ-tan2⁡θ1-tan2⁡2θtan2⁡θ=tan⁡3θtan⁡θOn taking left hand side,tan2⁡2θ+tan2⁡θ1-tan2⁡2θtan2⁡θ=(tan⁡2θ+tan⁡θ)(tan⁡2θ-tan⁡θ)(1-tan⁡2θtan⁡θ)(1+tan⁡2θtan⁡θ) (Now, a2-b2=(a-b)(a+b))It is known that,tan⁡(A+B)=tan⁡A+tan⁡B1-tan⁡Atan⁡Btan⁡(A-B)=tan⁡A-tan⁡B1+tan⁡Atan⁡BThus, the expression become,=tan⁡2θ+tan⁡θ1-tan⁡2θtan⁡θ×tan⁡2θ-tan⁡θ1+tan⁡2θtan⁡θ=tan⁡(2θ+θ)tan⁡(2θ-θ)=tan⁡3θtan⁡θSo,tan2⁡2θ-tan2⁡θ1-tan2⁡2θtan2⁡θ=tan⁡3θtan⁡θThus, the given relation is true.Option 1 is correct.