Question

Prove It - \[\tan \phi + 2\tan 2\phi + 4\tan 4\phi + 8\cot 8\phi = \cot \phi \]?

Answer 1

Hint: We need to prove \[\tan \phi + 2\tan 2\phi + 4\tan 4\phi + 8\cot 8\phi = \cot \phi \]. For this, we will mainly use the trigonometric identity \[\tan 2\theta = \dfrac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }}\]. Using this identity again and again from different angles, we will keep simplifying the given expression and then finally reach our answer. Also, we will be using \[\cot \theta = \dfrac{1}{{\tan \theta }}\].

Complete step-by-step answer:
We need to prove \[\tan \phi + 2\tan 2\phi + 4\tan 4\phi + 8\cot 8\phi = \cot \phi \].
Let us consider,
\[\tan \phi + 2\tan 2\phi + 4\tan 4\phi + 8\cot 8\phi \]
We know, \[\cot \theta = \dfrac{1}{{\tan \theta }}\]. So, we can write \[8\cot 8\phi \] as \[\dfrac{8}{{\tan 8\phi }}\]. Writing this, we get our left hand side to be equal to
\[ = \tan \phi + 2\tan 2\phi + 4\tan 4\phi + \dfrac{8}{{\tan 8\phi }}\]
Now, using \[\tan 2\theta = \dfrac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }}\] on the last term, we get
\[ = \tan \phi + 2\tan 2\phi + 4\tan 4\phi + \dfrac{8}{{\dfrac{{2\tan 4\phi }}{{1 - {{\tan }^2}4\phi }}}}\]
Now, using the property \[\dfrac{a}{{\dfrac{b}{c}}} = \dfrac{{ac}}{b}\], we get
\[ = \tan \phi + 2\tan 2\phi + 4\tan 4\phi + \dfrac{{8\left( {1 - {{\tan }^2}4\phi } \right)}}{{2\tan 4\phi }}\]
Now cancelling some term from numerator and denominator, we get
\[ = \tan \phi + 2\tan 2\phi + 4\tan 4\phi + \dfrac{{4\left( {1 - {{\tan }^2}4\phi } \right)}}{{\tan 4\phi }}\]
Now, taking LCM of the last two terms and opening the brackets, we get
\[ = \tan \phi + 2\tan 2\phi + \dfrac{{4{{\tan }^2}4\phi + 4 - 4{{\tan }^2}4\phi }}{{\tan 4\phi }}\]
Now, cancelling out \[4{\tan ^2}4\phi \] from the numerator, we get
\[ = \tan \phi + 2\tan 2\phi + \dfrac{4}{{\tan 4\phi }}\]
Again using \[\tan 2\theta = \dfrac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }}\] in the last term, we get
\[ = \tan \phi + 2\tan 2\phi + \dfrac{4}{{\dfrac{{2\tan 2\phi }}{{1 - {{\tan }^2}2\phi }}}}\]
Using the property \[\dfrac{a}{{\dfrac{b}{c}}} = \dfrac{{ac}}{b}\], we have
\[ = \tan \phi + 2\tan 2\phi + \dfrac{{4\left( {1 - {{\tan }^2}2\phi } \right)}}{{2\tan 2\phi }}\]
Cancelling some terms from the numerator and denominator, we get
\[ = \tan \phi + 2\tan 2\phi + \dfrac{{2\left( {1 - {{\tan }^2}2\phi } \right)}}{{\tan 2\phi }}\]
Now, taking LCM of the last two terms and opening the brackets, we get
\[ = \tan \phi + \dfrac{{2{{\tan }^2}2\phi + 2 - 2{{\tan }^2}2\phi }}{{\tan 2\phi }}\]
Now, cancelling \[2{\tan ^2}2\phi \] from the numerator, we get
\[ = \tan \phi + \dfrac{2}{{\tan 2\phi }}\]
Again using \[\tan 2\theta = \dfrac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }}\], we get our left hand side to be equal to
\[ = \tan \phi + \dfrac{2}{{\dfrac{{2\tan \phi }}{{1 - {{\tan }^2}\phi }}}}\]
Using \[\dfrac{a}{{\dfrac{b}{c}}} = \dfrac{{ac}}{b}\], we have
\[ = \tan \phi + \dfrac{{2\left( {1 - {{\tan }^2}\phi } \right)}}{{2\tan \phi }}\]
Now, cancelling out \[2\] from the numerator and denominator, we get
\[ = \tan \phi + \dfrac{{\left( {1 - {{\tan }^2}\phi } \right)}}{{\tan \phi }}\]
Now, taking LCM and opening the brackets, we get
\[ = \dfrac{{{{\tan }^2}\phi + 1 - {{\tan }^2}\phi }}{{\tan \phi }}\]
Cancelling \[{\tan ^2}\phi \] from the numerator, Left Hand Side becomes
\[ = \dfrac{1}{{\tan \phi }}\]
We know, \[\cot \phi = \dfrac{1}{{\tan \phi }}\]. So, using this Left Hand Side becomes
\[ \Rightarrow \dfrac{1}{{\tan \phi }} = \cot \phi \]
Hence, we get Left Hand Side = Right Hand Side \[ = \cot \phi \]

Note: We can also solve this problem using the formula \[\cot \phi - \tan \phi = 2\cot 2\phi \]. For that, we will first bring all our terms to the right hand side and then apply this formula first on the some terms according to the given conditions and then to the other terms and prove it equal to zero i.e. we need to prove \[RHS - LHS = 0\] which implies \[LHS = RHS\].