I needed to solve the following equation: tan ⁡ θ + tan ⁡ 2 θ + tan ⁡ 3 θ = tan ⁡ θ tan ⁡ 2 θ tan ⁡ 3 θTransform the LHS first: tan ⁡ θ + tan ⁡ 2 θ + tan ⁡ 3 θ = ( tan ⁡ θ + tan ⁡ 2 θ ) + tan ⁡ θ + tan ⁡ 2 θ 1 − tan ⁡ θ tan ⁡ 2 θ = ( tan ⁡ θ + tan ⁡ 2 θ ) ( 2 − tan ⁡ θ tan ⁡ 2 θ ) 1 − tan ⁡ θ tan ⁡ 2 θ And, RHS yields tan ⁡ θ tan ⁡ 2 θ tan ⁡ 3 θ = ( tan ⁡ θ tan ⁡ 2 θ ) tan ⁡ θ + tan ⁡ 2 θ 1 − tan ⁡ θ tan ⁡ 2 θ Now, two terms can be cancelled out from LHS and RHS, yielding the equation: 2 − tan ⁡ θ tan ⁡ 2 θ = tan ⁡ θ tan ⁡ 2 θ tan ⁡ θ tan ⁡ 2 θ = 1 , which can be further reduced as: tan 2 ⁡ θ = 1 3 ⟹ tan ⁡ θ = ± 1 3 Now, we can yield the general solution of this equation: θ = n π ± π 6 , n ∈ Z. But, setting θ = π 6 in the original equation is giving one term tan ⁡ π 2 , which is not defined.

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I needed to solve the following equation:  tan  ⁡  θ  +  tan  ⁡  2  θ  +  tan  ⁡  3  θ  =  tan  ⁡  θ  tan  ⁡  2  θ  tan  ⁡  3  θTransform the LHS first:                    tan        ⁡        θ        +        tan        ⁡        2        θ        +        tan        ⁡        3        θ                            =        (        tan        ⁡        θ        +        tan        ⁡        2        θ        )        +                                            tan              ⁡              θ              +              tan              ⁡              2              θ                                      1              −              tan              ⁡              θ              tan              ⁡              2              θ                                                                        =                                            (              tan              ⁡              θ              +              tan              ⁡              2              θ              )              (              2              −              tan              ⁡              θ              tan              ⁡              2              θ              )                                      1              −              tan              ⁡              θ              tan              ⁡              2              θ                                          And, RHS yields                    tan        ⁡        θ        tan        ⁡        2        θ        tan        ⁡        3        θ                            =        (        tan        ⁡        θ        tan        ⁡        2        θ        )                                            tan              ⁡              θ              +              tan              ⁡              2              θ                                      1              −              tan              ⁡              θ              tan              ⁡              2              θ                                          Now, two terms can be cancelled out from LHS and RHS, yielding the equation:                    2        −        tan        ⁡        θ        tan        ⁡        2        θ                            =        tan        ⁡        θ        tan        ⁡        2        θ                            tan        ⁡        θ        tan        ⁡        2        θ                            =        1        ,            which can be further reduced as:      tan    2    ⁡  θ  =      1    3      ⟹    tan  ⁡  θ  =  ±      1          3      Now, we can yield the general solution of this equation:  θ  =  n  π  ±            π      6        ,  n  ∈  Z. But, setting   θ  =            π      6       in the original equation is giving one term   tan  ⁡            π      2      , which is not defined.

Lisbonaid · Accepted Answer

When you cancel out the terms from LHS and RHS, you drop the solutions when these terms are 0 or do not exist (because the denominator is 0). A trivial example would the   θ  =  0, which certainly is a solution, but you did not find it because of the canceled terms.

opepayflarpws · Accepted Answer

You concluded from                    (        tan        ⁡        θ        +        tan        ⁡        2        θ        )        (        2        −        tan        ⁡        θ        tan        ⁡        2        θ        )                    1        −        tan        ⁡        θ        tan        ⁡        2        θ              =  (  tan  ⁡  θ  tan  ⁡  2  θ  )                    tan        ⁡        θ        +        tan        ⁡        2        θ                    1        −        tan        ⁡        θ        tan        ⁡        2        θ            (by “canceling”                     tan        ⁡        θ        +        tan        ⁡        2        θ                    1        −        tan        ⁡        θ        tan        ⁡        2        θ             from both sides) that  2  −  tan  ⁡  θ  tan  ⁡  2  θ  =  tan  ⁡  θ  tan  ⁡  2  θ  .The correct conclusion is that either                     tan        ⁡        θ        +        tan        ⁡        2        θ                    1        −        tan        ⁡        θ        tan        ⁡        2        θ              =  0 or   2  −  tan  ⁡  θ  tan  ⁡  2  θ  =  tan  ⁡  θ  tan  ⁡  2  θ And you still have to check your solutions and keep only those for which the original equation is defined.Canceling terms doesn’t give you an equivalent equation (one with the same solution set). If you cancel zero, you can lose solutions. If you cancel something undefined, you can introduce wrong solutions.For example, consider the equation      x          1      −      x        =      1          1      −      x        .If you “cancel” 1−x from each side, you get x=1, but that is not a solution, because the equation is not defined when x=1. And consider the equation      x    (      1    −    x    )  =  2  (      1    −    x    )  .If you “cancel” 1−x, the only solution to what’s left is x=2, and you lose the other solution (x=1) to the equation, because for that value, you canceled zero from each side.To summarize, if E⋅A=E⋅B, where E, A, and B are expressions, solutions occur when bothE⋅A and E⋅B are definedEither E=0 or A=B (or both).

I needed to solve the following equation: tan ⁡ θ + ta

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