Question

Prove that \[tan4x = \dfrac{{4tanx(1 - ta{n^2}x)}}{{(1 - 6ta{n^2}x + ta{n^4}x)}}\]

Answer 1

Hint:Here we write break the angle inside the bracket, \[4x = 2x + 2x\] and use the trigonometric formula of \[tan(A + B) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}\] to expand the value and then we use the formula of \[\tan 2x = \dfrac{{2\tan x}}{{1 - {{\tan }^2}x}}\] to solve further.

Complete step-by-step answer:
First we consider the LHS of the equation.
Since we know \[4x = 2x + 2x\], so we can write
\[tan{\text{ }}4x{\text{ }} = tan{\text{ }}\left( {2x{\text{ }} + 2x} \right)\]
Use the formula \[tan(A + B) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}\] and substitute the values of \[A = 2x,B = 2x\].
\[tan{\text{ }}4x{\text{ }} = \dfrac{{\tan 2x + \tan 2x}}{{1 - \tan 2x \times \tan 2x}}\]
Adding the terms in the numerator and multiplying the terms in the denominator we get
\[\tan 4x = \dfrac{{2\tan 2x}}{{1 - {{\tan }^2}2x}}\]
Now we use the formula \[\tan 2x = \dfrac{{2\tan x}}{{1 - {{\tan }^2}x}}\] and substitute in both numerator and denominator.
\[\tan 4x = \dfrac{{2\left[ {\dfrac{{2\tan x}}{{1 - {{\tan }^2}x}}} \right]}}{{1 - {{\left[ {\dfrac{{2\tan x}}{{1 - {{\tan }^2}x}}} \right]}^2}}}\]
\[
  \tan 4x = \dfrac{{\left[ {\dfrac{{2 \times 2\tan x}}{{1 - {{\tan }^2}x}}} \right]}}{{1 - \dfrac{{{{(2\tan x)}^2}}}{{{{(1 - {{\tan }^2}x)}^2}}}}} \\
  \tan 4x = \dfrac{{\left[ {\dfrac{{4\tan x}}{{1 - {{\tan }^2}x}}} \right]}}{{1 - \dfrac{{4{{\tan }^2}x}}{{{{(1 - {{\tan }^2}x)}^2}}}}} \\
 \]
Taking LCM in denominator.
\[
  \tan 4x = \dfrac{{\left[ {\dfrac{{4\tan x}}{{1 - {{\tan }^2}x}}} \right]}}{{\dfrac{{{{(1 - {{\tan }^2}x)}^2} - 4{{\tan }^2}x}}{{{{(1 - {{\tan }^2}x)}^2}}}}} \\
  \tan 4x = \dfrac{{4\tan x}}{{1 - {{\tan }^2}x}} \times \dfrac{{{{(1 - {{\tan }^2}x)}^2}}}{{{{(1 - {{\tan }^2}x)}^2} - 4{{\tan }^2}x}} \\
 \]
Using the formula \[{(a - b)^2} = {a^2} + {b^2} - 2ab\] substitute the value \[a = 1,b = {\tan ^2}x\]
\[{(1 - {\tan ^2}x)^2} = 1 + {({\tan ^2}x)^2} - 2{\tan ^2}x = 1 + {\tan ^4}x - 2{\tan ^2}x\]
\[
  \tan 4x = \dfrac{{4\tan x}}{{1 - {{\tan }^2}x}} \times \dfrac{{(1 - {{\tan }^2}x) \times (1 - {{\tan }^2}x)}}{{1 + {{\tan }^4}x - 2{{\tan }^2}x - 4{{\tan }^2}x}} \\
  \tan 4x = \dfrac{{4\tan x}}{{1 - {{\tan }^2}x}} \times \dfrac{{(1 - {{\tan }^2}x) \times (1 - {{\tan }^2}x)}}{{1 + {{\tan }^4}x - 6{{\tan }^2}x}} \\
 \]
Cancel out the same terms from numerator and denominator.
\[\tan 4x = \dfrac{{4\tan x(1 - {{\tan }^2}x)}}{{1 + {{\tan }^4}x - 6{{\tan }^2}x}}\]
Therefore reshuffling the terms in the denominator we can write
\[\tan 4x = \dfrac{{4\tan x(1 - {{\tan }^2}x)}}{{1 - 6{{\tan }^2}x + {{\tan }^4}x}}\]
After solving we get LHS of the equation equal to RHS of the equation.
Hence proved.

Note:Students are likely to make mistake if they substitute the formula direct into the next step as here the angle is not \[x\] but it is \[2x\] which is a double angle, so it is recommended to substitute the values into the formula first and then substitute the values into the next step. Students should avoid solving this question by substitution of \[\tan 4x = \dfrac{{\sin 4x}}{{\cos 4x}}\] because then it will be very complex to solve as we will have to substitute for both the values of numerator and denominator as in the RHS the angle is \[x\] so we will have to substitute in the formulas until we come to the angle \[x\].
Also, students should not break the value of \[4x = x + 3x\] because then we will have to use a formula for \[\tan 3x\] which will be complex.