Question

Show that tan 3x tan 2x tan x = tan 3x – tan 2x – tan x.



No spam​

Answer 1

$\\\;\underbrace{\underline{\sf{Understanding\;the\;Concept}}}$

Here the concept of Trigonometry has been used. This question seems of secondary grade There are number of ways to solve this. But we shall be using the appropriate method to solve this question. Firstly we have to make relation between the terms of x. Then we will apply tan function there and then find our answer.

Let's do it !!

______________________________________________

★ Formula Used :-

$\\\;\boxed{\sf{\pink{\tan\:(A\;+\;B)\;=\;\bf{\dfrac{tan\:A\;+\;\tan\:B}{1\;-\;\tan\:A\;\tan\:B}}}}}$

______________________________________________

To Prove :-

$\\\;\bf{\mapsto\;\;\green{\tan\:3x\;\tan\:2x\;\tan\:x\;=\;\tan\:3x\;-\;\tan\:2x\;-\;\tan\:x}}$

______________________________________________

★ Solution :-

We are already given the equation. Firstly, we shall find the relation between x.

We know that ,

$\\\;\bf{\rightarrow\;\;\blue{3\;=\;\bf{2\;+\;1}}}$

Then multiplying x to all the terms, we get

$\\\;\bf{\rightarrow\;\;3x\;=\;\bf{x(2\;+\;1)}}$

$\\\;\bf{\rightarrow\;\;3x\;=\;\bf{2x\;+\;x}}$

Now multiplying all the terms by tan, we get

$\\\;\bf{\rightarrow\;\;\tan\;3x\;=\;\bf{\tan\:(2x\;+\;x)}}$

By using the formula, we get

$\\\;\sf{\tan\:(A\;+\;B)\;=\;\bf{\dfrac{tan\:A\;+\;\tan\:B}{1\;-\;\tan\:A\;\tan\:B}}}$

• Here A = 2x and B = x

Let's now apply this formula in the equation we got. Then,

$\\\;\bf{\Longrightarrow\;\;\tan\;3x\;=\;\bf{\orange{\dfrac{\tan\:2x\;+\;\tan\:x}{1\;-\;\tan\:2x\;\tan\:x}}}}$

Now cross multiplying the terms, we get

$\\\;\bf{\Longrightarrow\;\;\tan\;3x\:\times\:(1\;-\;\tan\:2x\;\tan\:x)\;=\;\bf{\tan\:2x\;+\;\tan\:x}}$

$\\\;\bf{\Longrightarrow\;\;\tan\;3x\;-\;\tan\:3x\;\tan\:2x\;\tan\:x\;=\;\bf{\tan\:2x\;+\;\tan\:x}}$

Now transposing the like terms, we get

$\\\;\bf{\Longrightarrow\;\;\tan\;3x\;-\;\tan\:2x\;-\;\tan\:x\;=\;\bf{\tan\:3x\;\tan\:2x\;\tan\:x}}$

$\\\;\bf{\Longrightarrow\;\;\red{\tan\:3x\;\tan\:2x\;\tan\:x\;=\;\tan\;3x\;-\;\tan\:2x\;-\;\tan\:x\;}}$

This is the thing to be proved. So we got our answer.

$\\\;\qquad\qquad\boxed{\underline{\tt{\purple{Hence,\;\;Proved}}}}$

______________________________________________

★ More to know :-

$\\\;\sf{\leadsto\;\;\sin^{2}\:\theta\;=\;1\;-\;\cos^{2}\:\theta}$

$\\\;\sf{\leadsto\;\;cosec^{2}\:\theta\;=\;1\;+\;\tan^{2}\:\theta}$

$\\\;\sf{\leadsto\;\;\sec^{2}\:\theta\;=\;1\;+\;\cot^{2}\:\theta}$

$\\\;\sf{\leadsto\;\;\sin\theta\;=\;\cos(90^{\circ}\;-\;\theta)}$

$\\\;\sf{\leadsto\;\;cosec\theta\;=\;\sec(90^{\circ}\;-\;\theta)}$

$\\\;\sf{\leadsto\;\;\tan\theta\;=\;\cot(90^{\circ}\;-\;\theta)}$

Answer 2

$\large \: \underline{ \rm \color{navy} \: SolutiOn :- }$

tan ( 2x + x ). Because in question it is in term of 2x - x. By applying this we can easily prove our question.

Let us take ,$\rm \: 3x = \tan(2x + x)$

Now we can apply of tan ( A + B ) i.e.

$\\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \blue{ \underline{\sf \boxed{ \: \green{\sf\tan(A + B) = \frac{ \tan \: A + \tan B }{1 - \tan\: A \: \tan B} }}}}$

Now, We can write,

$\\ \: \: \: \sf \:tan 3x = tan (2x+x) = \frac{ \tan \: A + \tan B }{1 - \tan\: A \: \tan B}$

Now we cross multiple to get ,

$\\ \: \: \: \: \: \: \: \: \: \: \: \: \longrightarrow \sf \: \: tan( 3x)[ 1 - tan(2x )\: tan(x)] \: = \: tan(2x) + tan(x)$

$\\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \longrightarrow \: \sf \: tan(3x) - tan(3x) \: tan(2x) \: tan(x) \: = \: tan(2x) + tan(x)$

Now, Rearranging the above equation we get ,

$\\ \sf \: \: \: \: \: \: \longrightarrow \: tan(3x) - tan(2x) - tan(x) = tan(3x) \: tan(2x) \: tan(x)$

$\\ \: \: \: \: \: \: \: \: \: \underline{ \rm \green {\fbox{{Hence, Proved}}}}$

$\large \: \underline{ \rm \color{navy} \: Note :- }$

Whenever we face such type of questions the key concept for solving the question is that we always try to make the bigger angle in sum of 2 small angles that are given in question to apply the formula to prove the question.