Answer

Verified

417k+ views

**Hint:**These types of problems are pretty straight forward and are very simple to solve. Problems like these require prior knowledge of trigonometric equations and other various trigonometric formulae. The main catch of the problem is to multiply the numerator and denominator of each and every term in the left hand side by a factor and then to proceed further with it and convert it then into the required tangent form by multiplying and dividing \[2\] on the numerator and denominator.

**Complete step by step answer:**

Now, we start off the solution to the given problem as,

We consider the first term of the left hand side, \[\dfrac{\sin x}{cos3x}\] . We multiply and divide the numerator and the denominator with \[\cos x\] , it then transforms to,

\[\dfrac{\sin x\cos x}{cos3x\cos x}\] . Now, multiplying and dividing the numerator and denominator by \[2\] we get,

\[\begin{align}

& \dfrac{1}{2}\dfrac{2\sin x\cos x}{cos3x\cos x} \\

& \Rightarrow \dfrac{1}{2}\dfrac{\sin 2x}{cos3x\cos x} \\

& \Rightarrow \dfrac{1}{2}\dfrac{\sin \left( 3x-x \right)}{cos3x\cos x} \\

& \Rightarrow \dfrac{1}{2}\dfrac{\sin 3x\cos x-\cos 3x\sin x}{cos3x\cos x} \\

& \Rightarrow \dfrac{1}{2}\left[ \dfrac{\sin 3x\cos x}{cos3x\cos x}-\dfrac{\cos 3x\sin x}{cos3x\cos x} \right] \\

& \Rightarrow \dfrac{1}{2}\left[ \dfrac{\sin 3x}{cos3x}-\dfrac{\sin x}{\cos x} \right] \\

& \Rightarrow \dfrac{1}{2}\left[ \tan 3x-\tan x \right] \\

\end{align}\]

Now, simplifying the second term by following the same steps we get,

\[\begin{align}

& \dfrac{1}{2}\dfrac{2\sin 3x\cos 3x}{cos9x\cos 3x} \\

& \Rightarrow \dfrac{1}{2}\dfrac{\sin 6x}{cos9x\cos 3x} \\

& \Rightarrow \dfrac{1}{2}\dfrac{\sin \left( 9x-3x \right)}{cos9x\cos 3x} \\

& \Rightarrow \dfrac{1}{2}\dfrac{\sin 9x\cos 3x-\cos 9x\sin 3x}{cos9x\cos 3x} \\

& \Rightarrow \dfrac{1}{2}\left[ \dfrac{\sin 9x\cos 3x}{cos9x\cos 3x}-\dfrac{\cos 9x\sin 3x}{cos9x\cos 3x} \right] \\

& \Rightarrow \dfrac{1}{2}\left[ \dfrac{\sin 9x}{cos9x}-\dfrac{\sin 3x}{\cos 3x} \right] \\

& \Rightarrow \dfrac{1}{2}\left[ \tan 9x-\tan 3x \right] \\

\end{align}\]

Now, simplifying the third term by following the same steps we get,

\[\begin{align}

& \dfrac{1}{2}\dfrac{2\sin 9x\cos 9x}{cos27x\cos 9x} \\

& \Rightarrow \dfrac{1}{2}\dfrac{\sin 18x}{cos27x\cos 9x} \\

& \Rightarrow \dfrac{1}{2}\dfrac{\sin \left( 27x-9x \right)}{cos27x\cos 9x} \\

& \Rightarrow \dfrac{1}{2}\dfrac{\sin 27x\cos 9x-\cos 27x\sin 9x}{cos27x\cos 9x} \\

& \Rightarrow \dfrac{1}{2}\left[ \dfrac{\sin 27x\cos 9x}{cos27x\cos 9x}-\dfrac{\cos 27x\sin 9x}{cos27x\cos 9x} \right] \\

& \Rightarrow \dfrac{1}{2}\left[ \dfrac{\sin 27x}{cos27x}-\dfrac{\sin 9x}{\cos 9x} \right] \\

& \Rightarrow \dfrac{1}{2}\left[ \tan 27x-\tan 9x \right] \\

\end{align}\]

Now adding all the three terms in their simplified form we get,

\[\dfrac{1}{2}\left[ \tan 3x-\tan x \right]+\dfrac{1}{2}\left[ \tan 9x-\tan 3x \right]+\dfrac{1}{2}\left[ \tan 27x-\tan 9x \right]\]

Adding up all these we get,

\[\begin{align}

& \Rightarrow \dfrac{1}{2}\left[ \tan 3x-\tan x+\tan 9x-\tan 3x+\tan 27x-\tan 9x \right] \\

& \Rightarrow \dfrac{1}{2}\left[ \tan 27x-\tan x \right] \\

\end{align}\]

Thus, it gets evaluated to the above value. Now if we take a closer look, we see that the equation that we have arrived at is the same as that written in the right hand side of the given problem. Thus we say that we have proved the required equation correctly.

**Note:**For problems like these, we need to recall all the formulae and equations of trigonometry. The main catch of this problem is the multiplication of a term in both the numerator and denominator to simplify our problem. This idea of multiplication of a common term, will come only by more and more practice, or else it would be very difficult for us to solve these types of problems.

Recently Updated Pages

what is the correct chronological order of the following class 10 social science CBSE

Which of the following was not the actual cause for class 10 social science CBSE

Which of the following statements is not correct A class 10 social science CBSE

Which of the following leaders was not present in the class 10 social science CBSE

Garampani Sanctuary is located at A Diphu Assam B Gangtok class 10 social science CBSE

Which one of the following places is not covered by class 10 social science CBSE

Trending doubts

A rainbow has circular shape because A The earth is class 11 physics CBSE

Which are the Top 10 Largest Countries of the World?

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

How do you graph the function fx 4x class 9 maths CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

Why is there a time difference of about 5 hours between class 10 social science CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Difference Between Plant Cell and Animal Cell