Question

Prove that $\tan 4x=\dfrac{4\tan x\left( 1-{{\tan }^{2}}x \right)}{1+{{\tan }^{4}}x-6{{\tan }^{2}}x}$

Answer 1

Hint: Use the identity $\tan \left( 2x \right)=\dfrac{2\tan x}{1-{{\tan }^{2}}x}$. Write tan(4x) = tan (2(2x)) and apply the formula. Then apply the formula again and simplify to get the result. Use algebraic identity ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$.
Alternatively, you can use $\tan \left( nx \right)=\dfrac{^{n}{{C}_{1}}\tan x{{-}^{n}}{{C}_{3}}{{\tan }^{3}}x{{+}^{n}}{{C}_{5}}{{\tan }^{5}}x+\cdots }{^{n}{{C}_{0}}{{-}^{n}}{{C}_{2}}{{\tan }^{2}}x{{+}^{n}}{{C}_{4}}{{\tan }^{4}}x+\cdots }$
Put n = 4 in the above formula to get the result.
Alternatively, expand the expression ${{\left( \cos x+i\sin x \right)}^{4}}$ and use De Movire’s formula and equal real and imaginary parts and hence find the expression for tan4x

Complete step-by-step answer:
We have tan(4x) = tan(2(2x))
Using $\tan \left( 2x \right)=\dfrac{2\tan x}{1-{{\tan }^{2}}x}$, we get
$\tan 4x=\dfrac{2\tan 2x}{1-{{\tan }^{2}}2x}$
We know that $\tan \left( 2x \right)=\dfrac{2\tan x}{1-{{\tan }^{2}}x}$.
Using, we get
$\tan 4x=\dfrac{2\left( \dfrac{2\tan x}{1-{{\tan }^{2}}x} \right)}{1-{{\left( \dfrac{2\tan x}{1-{{\tan }^{2}}x} \right)}^{2}}}$
Multiplying the numerator and denominator by ${{\left( 1-{{\tan }^{2}}x \right)}^{2}}$, we get
$\tan 4x=\dfrac{2\left( 2\tan x \right)\left( 1-{{\tan }^{2}}x \right)}{{{\left( 1-{{\tan }^{2}}x \right)}^{2}}-4{{\tan }^{2}}x}$
Using ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$, we get
$\tan 4x=\dfrac{4\tan x\left( 1-{{\tan }^{2}}x \right)}{1+{{\tan }^{4}}x-2{{\tan }^{2}}x-4{{\tan }^{2}}x}=\dfrac{4\tan x\left( 1-{{\tan }^{2}}x \right)}{1+{{\tan }^{4}}x-6{{\tan }^{2}}x}$
Hence LHS = RHS and hence the result is proved.

Note: [1] The formulae $\sin 2x=\dfrac{2\tan x}{1+{{\tan }^{2}}x},\cos 2x=\dfrac{1-{{\tan }^{2}}x}{1+{{\tan }^{2}}x}$ and $\tan \left( 2x \right)=\dfrac{2\tan x}{1-{{\tan }^{2}}x}$ are called double angle formulae.
[2] Alternative solution: Best Method
We know $\tan \left( nx \right)=\dfrac{^{n}{{C}_{1}}\tan x{{-}^{n}}{{C}_{3}}{{\tan }^{3}}x{{+}^{n}}{{C}_{5}}{{\tan }^{5}}x+\cdots }{^{n}{{C}_{0}}{{-}^{n}}{{C}_{2}}{{\tan }^{2}}x{{+}^{n}}{{C}_{4}}{{\tan }^{4}}x+\cdots }$
Now $^{4}{{C}_{0}}=1{{,}^{4}}{{C}_{1}}=4{{,}^{4}}{{C}_{2}}=6{{,}^{4}}{{C}_{3}}=4$ and $^{4}{{C}_{4}}=1$
Hence, we have
$\tan 4x=\dfrac{4\tan x-4{{\tan }^{3}}x}{1-6{{\tan }^{2}}x+{{\tan }^{4}}x}$
Hence LHS = RHS.
[3] The above result can be obtained by applying binomial theorem to the expression ${{\left( \cos x+i\sin x \right)}^{n}}$ and using De Moivre's identity ${{\left( \cos x+i\sin x \right)}^{n}}=\cos nx+i\sin nx$.
Compare Real and imaginary parts and use $\tan nx=\dfrac{\sin nx}{\cos nx}$.
[4] Alternatively, we have
${{\left( \cos x+i\sin x \right)}^{4}}=\cos 4x+i\sin 4x$.
We know that ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$.
Using the above formula, we get
$\begin{aligned} & {{\left( {{\left( \cos x+i\sin x \right)}^{2}} \right)}^{2}}=\cos 4x+i\sin 4x \\\ & \Rightarrow {{\left( {{\cos }^{2}}x-{{\sin }^{2}}x+2i\sin x\cos x \right)}^{2}}=\cos 4x+i\sin 4x \\\ \end{aligned}$
We know that ${{\left( a+b+c \right)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2ab+2bc+2ac$
Using the above identity, we get
$\begin{aligned} & {{\left( {{\cos }^{2}}x-{{\sin }^{2}}x+2i\sin x\cos x \right)}^{2}}=\cos 4x+i\sin 4x \\\ & \Rightarrow {{\cos }^{4}}x+{{\sin }^{4}}x-4{{\sin }^{2}}x{{\cos }^{2}}x-2{{\cos }^{2}}x{{\sin }^{2}}x+4i\sin x{{\cos }^{3}}x-4i{{\sin }^{3}}x\cos x=\cos 4x+i\sin 4x \\\ \end{aligned}$
Comparing Real and imaginary parts, we have
$\cos 4x={{\cos }^{4}}x+{{\sin }^{4}}x-6{{\sin }^{2}}x{{\cos }^{2}}x$ and $\sin 4x=\sin x{{\cos }^{3}}x-\cos x{{\sin }^{3}}x$
Hence, $\tan 4x=\dfrac{4\sin x{{\cos }^{3}}x-4\cos x{{\sin }^{3}}x}{{{\cos }^{4}}x+{{\sin }^{4}}x-6{{\sin }^{2}}x{{\cos }^{2}}x}$
Dividing numerator and denominator by ${{\cos }^{4}}x$, we get
$\tan 4x=\dfrac{4\tan x-4{{\tan }^{3}}x}{1-6{{\tan }^{2}}x+{{\tan }^{4}}x}$
Hence LHS = RHS and hence the result is proved.