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Question: Prove that \( \tan 3\theta = \dfrac{{3\tan \theta - {{\tan }^3}\theta }}{{1 - 3{{\tan }^2}\theta }} ...

Prove that tan⁡3θ=3tan⁡θ−tan⁡3θ1−3tan⁡2θ\tan 3\theta = \dfrac{{3\tan \theta - {{\tan }^3}\theta }}{{1 - 3{{\tan }^2}\theta }} .

Explanation

Solution

Hint : The given question deals with basic simplification of trigonometric functions by using some of the simple trigonometric formulae such as the compound angle formula of tangent, tan⁡(A+B)=(tan⁡A+tan⁡B1−tan⁡Atan⁡B)\tan \left( {A + B} \right) = \left( {\dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}} \right) . Basic algebraic rules and trigonometric identities are to be kept in mind while simplifying the given problem and proving the result given to us.

Complete step-by-step answer :
In the given problem, we have to prove a trigonometric equality that can be further used in many questions and problems as a direct result and has wide ranging applications. For proving the desired result, we need to have a good grip over the basic trigonometric formulae and identities.
Now, we need to make the left and right sides of the equation equal.
L.H.S. =tan⁡3θ= \tan 3\theta
So, we first break down the angle in the tangent function into two parts.
=tan⁡(2θ+θ)= \tan \left( {2\theta + \theta } \right)
Now, we have to apply the compound angle formula for tangent trigonometric function so as to simplify the expression and prove the equality. So, we have the trigonometric formula tan⁡(A+B)=(tan⁡A+tan⁡B1−tan⁡Atan⁡B)\tan \left( {A + B} \right) = \left( {\dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}} \right) .
=(tan⁡2θ+tan⁡θ1−tan⁡2θ×tan⁡θ)= \left( {\dfrac{{\tan 2\theta + \tan \theta }}{{1 - \tan 2\theta \times \tan \theta }}} \right)
Now, we will use the expression of the double angle formula for tangent function. So, we have, tan⁡2x=(2tan⁡x1−tan⁡2x)\tan 2x = \left( {\dfrac{{2\tan x}}{{1 - {{\tan }^2}x}}} \right) . Now, substituting this, we get,
=(2tan⁡θ1−tan⁡2θ+tan⁡θ1−2tan⁡θ1−tan⁡2θ×tan⁡θ)= \left( {\dfrac{{\dfrac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }} + \tan \theta }}{{1 - \dfrac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }} \times \tan \theta }}} \right)
Now, taking the LCM of the rational expressions, we get,
=(2tan⁡θ+tan⁡θ(1−tan⁡2θ)1−tan⁡2θ(1−tan⁡2θ)−2tan⁡2θ1−tan⁡2θ)= \left( {\dfrac{{\dfrac{{2\tan \theta + \tan \theta \left( {1 - {{\tan }^2}\theta } \right)}}{{1 - {{\tan }^2}\theta }}}}{{\dfrac{{\left( {1 - {{\tan }^2}\theta } \right) - 2{{\tan }^2}\theta }}{{1 - {{\tan }^2}\theta }}}}} \right)
Cancelling the common factors in numerator and denominator, we get,
=(2tan⁡θ+tan⁡θ(1−tan⁡2θ)(1−tan⁡2θ)−2tan⁡2θ)= \left( {\dfrac{{2\tan \theta + \tan \theta \left( {1 - {{\tan }^2}\theta } \right)}}{{\left( {1 - {{\tan }^2}\theta } \right) - 2{{\tan }^2}\theta }}} \right)
Opening the brackets, we get,
=(2tan⁡θ+tan⁡θ−tan⁡3θ1−tan⁡2θ−2tan⁡2θ)= \left( {\dfrac{{2\tan \theta + \tan \theta - {{\tan }^3}\theta }}{{1 - {{\tan }^2}\theta - 2{{\tan }^2}\theta }}} \right)
Now, simplifying the expression, we get,
=(3tan⁡θ−tan⁡3θ1−3tan⁡2θ)= \left( {\dfrac{{3\tan \theta - {{\tan }^3}\theta }}{{1 - 3{{\tan }^2}\theta }}} \right)
Now, L.H.S =(3tan⁡θ−tan⁡3θ1−3tan⁡2θ)= \left( {\dfrac{{3\tan \theta - {{\tan }^3}\theta }}{{1 - 3{{\tan }^2}\theta }}} \right)
As the left side of the equation is equal to the right side of the equation, we have,
tan⁡3θ=3tan⁡θ−tan⁡3θ1−3tan⁡2θ\tan 3\theta = \dfrac{{3\tan \theta - {{\tan }^3}\theta }}{{1 - 3{{\tan }^2}\theta }}
Hence, Proved.

Note : Given problem deals with Trigonometric functions. For solving such problems, trigonometric formulae should be remembered by heart. Besides these simple trigonometric formulae, trigonometric identities are also of significant use in such type of questions where we have to simplify trigonometric expressions with help of basic knowledge of algebraic rules and operations. We must remember the compound angle formula and double angle formula of tangent function in order to solve the problem. One must take care while handling the calculations.