# Prove the following trigonometric equation.

$\tan \theta + \tan (60 + \theta ) + \tan (120 + \theta ) = 3\tan 3\theta $

Last updated date: 20th Mar 2023

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Answer

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Hint- For solving uses the identities of trigonometry. Take a single term at a time. Try to put the values of various trigonometric terms whose values are known and solve.

Complete step-by-step answer:

Given equation:

$ \Rightarrow \tan \theta + \tan (60 + \theta ) + \tan (120 + \theta ) = 3\tan 3\theta $

By taking L.H.S of the equation, we will proceed further

We know that

\[\left[ {\tan (A + B) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}} \right]\]

So substituting in the given equation we have

$ = \tan \theta + \dfrac{{\tan {{60}^0} + \tan \theta }}{{1 - \tan {{60}^0}\tan \theta }} + \dfrac{{\tan {{120}^0} + \tan \theta }}{{1 - \tan {{120}^0}\tan \theta }}$

Put the value of $\tan {60^0} = \sqrt 3 $ and $\tan {120^0} = \tan ({90^0} + {30^0}) = - \sqrt 3 $

$ = \tan \theta + \dfrac{{\sqrt 3 + \tan \theta }}{{1 - \sqrt 3 \tan \theta }} + \dfrac{{ - \sqrt 3 + \tan \theta }}{{1 - ( - \sqrt 3 )\tan \theta }}$

By taking LCM, we will proceed further

$

= \dfrac{{\tan \theta (1 - 3{{\tan }^2}\theta ) + (\sqrt 3 + \tan \theta )(1 + \sqrt 3 \tan \theta ) + ( - \sqrt 3 + \tan \theta )(1 - \sqrt 3 \tan \theta )}}{{(1 - \sqrt 3 \tan \theta )(1 + \sqrt 3 \tan \theta )}} \\

= \dfrac{{(\tan \theta - 3{{\tan }^3}\theta ) + (\sqrt 3 + 3\tan \theta + \tan \theta + \sqrt 3 {{\tan }^2}\theta ) + ( - \sqrt 3 + 3\tan \theta + \tan \theta - \sqrt 3 {{\tan }^2}\theta )}}{{(1 - 3{{\tan }^2}\theta )}} \\

$

As we know that $[(a - b)(a + b) = {a^2} - {b^2}]$

$ = \dfrac{{\tan \theta - 3{{\tan }^3}\theta + \sqrt 3 + 3\tan \theta + \tan \theta + \sqrt 3 {{\tan }^2}\theta - \sqrt 3 + 3\tan \theta + \tan \theta - \sqrt 3 {{\tan }^2}\theta }}{{1 - 3{{\tan }^2}\theta }}$

After cancelling the like terms we get

$

= \dfrac{{9\tan \theta - 3{{\tan }^3}\theta }}{{1 - 3{{\tan }^2}\theta }} \\

= \dfrac{{3(3\tan \theta - {{\tan }^3}\theta )}}{{1 - 3{{\tan }^2}\theta }} \\

$

As we know that $ \Rightarrow \dfrac{{3\tan \theta - {{\tan }^3}\theta }}{{1 - 3{{\tan }^2}\theta }} = \tan 3\theta $

$ \Rightarrow 3\tan 3A = $ R.H.S

Hence the given equation is proved.

Note- For solving such types of questions which involve complex trigonometric terms, solve the equations by the use of trigonometric identities keeping in mind the value to be proved. In order to find higher values of trigonometric angles such as $\tan {120^0}$ , in this case try to simplify the angle in multiple of ${90^0}$ and some angle.

Complete step-by-step answer:

Given equation:

$ \Rightarrow \tan \theta + \tan (60 + \theta ) + \tan (120 + \theta ) = 3\tan 3\theta $

By taking L.H.S of the equation, we will proceed further

We know that

\[\left[ {\tan (A + B) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}} \right]\]

So substituting in the given equation we have

$ = \tan \theta + \dfrac{{\tan {{60}^0} + \tan \theta }}{{1 - \tan {{60}^0}\tan \theta }} + \dfrac{{\tan {{120}^0} + \tan \theta }}{{1 - \tan {{120}^0}\tan \theta }}$

Put the value of $\tan {60^0} = \sqrt 3 $ and $\tan {120^0} = \tan ({90^0} + {30^0}) = - \sqrt 3 $

$ = \tan \theta + \dfrac{{\sqrt 3 + \tan \theta }}{{1 - \sqrt 3 \tan \theta }} + \dfrac{{ - \sqrt 3 + \tan \theta }}{{1 - ( - \sqrt 3 )\tan \theta }}$

By taking LCM, we will proceed further

$

= \dfrac{{\tan \theta (1 - 3{{\tan }^2}\theta ) + (\sqrt 3 + \tan \theta )(1 + \sqrt 3 \tan \theta ) + ( - \sqrt 3 + \tan \theta )(1 - \sqrt 3 \tan \theta )}}{{(1 - \sqrt 3 \tan \theta )(1 + \sqrt 3 \tan \theta )}} \\

= \dfrac{{(\tan \theta - 3{{\tan }^3}\theta ) + (\sqrt 3 + 3\tan \theta + \tan \theta + \sqrt 3 {{\tan }^2}\theta ) + ( - \sqrt 3 + 3\tan \theta + \tan \theta - \sqrt 3 {{\tan }^2}\theta )}}{{(1 - 3{{\tan }^2}\theta )}} \\

$

As we know that $[(a - b)(a + b) = {a^2} - {b^2}]$

$ = \dfrac{{\tan \theta - 3{{\tan }^3}\theta + \sqrt 3 + 3\tan \theta + \tan \theta + \sqrt 3 {{\tan }^2}\theta - \sqrt 3 + 3\tan \theta + \tan \theta - \sqrt 3 {{\tan }^2}\theta }}{{1 - 3{{\tan }^2}\theta }}$

After cancelling the like terms we get

$

= \dfrac{{9\tan \theta - 3{{\tan }^3}\theta }}{{1 - 3{{\tan }^2}\theta }} \\

= \dfrac{{3(3\tan \theta - {{\tan }^3}\theta )}}{{1 - 3{{\tan }^2}\theta }} \\

$

As we know that $ \Rightarrow \dfrac{{3\tan \theta - {{\tan }^3}\theta }}{{1 - 3{{\tan }^2}\theta }} = \tan 3\theta $

$ \Rightarrow 3\tan 3A = $ R.H.S

Hence the given equation is proved.

Note- For solving such types of questions which involve complex trigonometric terms, solve the equations by the use of trigonometric identities keeping in mind the value to be proved. In order to find higher values of trigonometric angles such as $\tan {120^0}$ , in this case try to simplify the angle in multiple of ${90^0}$ and some angle.

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