prove that tan 3a tan 2a tan a = tan 3a − tan 2a − tan a


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Question Number 38495 by kunal1234523 last updated on 26/Jun/18

$p r o v e t h a t$ $\tan 3 a \tan 2 a \tan a = \tan 3 a - \tan 2 a - \tan a$
	Answered by kunal1234523 last updated on 26/Jun/18

	$t a n 3 a = t a n (a + 2 a)$ $\Rightarrow t a n 3 a = \frac{t a n a + t a n 2 a}{1 - t a n a t a n 2 a}$ $\Rightarrow t a n 3 a - t a n a t a n 2 a t a n 3 a = t a n a + t a n 2 a$ $\Rightarrow t a n 3 a t a n 2 a t a n a = t a n 3 a - t a n 2 a - t a n a$
	Answered by kunal1234523 last updated on 26/Jun/18

	$t a n 2 a = t a n (3 a - a)$ $\Rightarrow t a n 2 a = \frac{t a n 3 a - t a n a}{1 + t a n 3 a t a n a}$ $\Rightarrow t a n 2 a + t a n 3 a t a n 2 a t a n a = t a n 3 a - t a n a$ $\Rightarrow t a n 3 a t a n 2 a t a n a = t a n 3 a - t a n 2 a - t a n a$ $also$ $t a n a = t a n (3 a - 2 a)$ $\Rightarrow t a n a = \frac{t a n 3 a - t a n 2 a}{1 + t a n 3 a t a n 2 a}$ $\Rightarrow t a n a + t a n 3 a t a n 2 a t a n a = t a n 3 a - t a n 2 a$ $\Rightarrow t a n 3 a t a n 2 a t a n a = t a n 3 a - t a n 2 a - t a n a$
	Answered by kunal1234523 last updated on 26/Jun/18

	$b u t t h e r e s h o u l d b e o n e m o r e a p p o r a c h a n d$ $I a m s t u c k e d t h e r e$ $L H S$ $\frac{s i n 3 a s i n 2 a s i n a}{c o s 3 a c o s 2 a c o s a}$ $= \frac{s i n 3 a (c o s a - c o s 3 a)}{2 c o s 3 a c o s 2 a c o s a}$ $= \frac{s i n 3 a c o s a}{2 c o s 3 a c o s 2 a c o s a} - \frac{s i n 3 a c o s 3 a}{2 c o s 3 a c o s 2 a c o s a}$ $= \frac{t a n 3 a}{2 c o s 2 a} - \frac{s i n (a + 2 a)}{2 c o s a c o s 2 a}$ $= \frac{t a n 3 a}{2 c o s 2 a} - (\frac{s i n a c o s 2 a}{2 c o s a c o s 2 a} + \frac{c o s a s i n 2 a}{2 c o s a c o s 2 a})$ $= \frac{t a n 3 a}{2 c o s 2 a} - \frac{t a n 2 a}{2} - \frac{t a n a}{2}$ $= \frac{1}{2} [\frac{t a n 3 a}{c o s 2 a} - t a n 2 a - t a n a]$ $n o w w h a t s h o u l d I d o . . . . .$
	Commented by tanmay.chaudhury50@gmail.com last updated on 26/Jun/18

	$s c a n n i n g i t m e t i c u l o u s l y . . . g i v e t i m e . . .$
	Commented by tanmay.chaudhury50@gmail.com last updated on 26/Jun/18

	$\frac{t a n 3 a}{2 c o s 2 a} - \frac{t a n 2 a}{2} - \frac{t a n a}{2}$ $t a n 3 a - t a n 2 a - t a n a + \frac{t a n 3 a}{2 c o s 2 a} - t a n 3 a + t a n 2 a$ $- \frac{t a n 2 a}{2} + t a n a - \frac{t a n a}{2}$ $t a n 3 a - t a n 2 a - t a n a + t a n 3 a (\frac{1}{2 c o s 2 a} - 1) + \frac{1}{2}$ $(t a n 2 a + t a n a)$ $d o + t a n 3 a (\frac{1}{2 c o s 2 a} - 1) + \frac{1}{2} (\frac{s i n 3 a}{c o s 2 a c o s a})$ $= d o + \frac{s i n 3 a}{2 c o s 3 a c o s 2 a} - \frac{s i n 3 a}{c o s 3 a} + \frac{s i n 3 a}{2 c o s 2 a c o s a}$ $d o + s i n 3 a (\frac{1}{2 c o s 3 a c o s 2 a} - \frac{1}{c o s 3 a} + \frac{1}{2 c o s 2 a c o s a})$ $d o + s i n 3 a (\frac{c o s a + c o s 3 a}{2 c o s 3 a c o s 2 a c o s a} - \frac{1}{c o s 3 a})$ $d o + s i n 3 a (\frac{2 c o s 2 a . c o s a}{2 c o s 3 a c o s 2 a c o s a} - \frac{1}{c o s 3 a})$ $d o + s i n 3 a (\frac{1}{c o s 3 a} - \frac{1}{c o s 3 a})$ $= d o + 0$ $= t a n 3 a - t a n 2 a - t a n a p r o v e d$
	Commented by kunal1234523 last updated on 27/Jun/18

	$w o w y o u t a k e a ‘ ‘ {d o}^{″} t h e n p r o v e d e v e r y t h i n g$ $e l s e i s z e r o s m a r t .$
	Answered by MJS last updated on 26/Jun/18

	$1 = \frac{\tan 3 α - \tan 2 α - \tan α}{\tan 3 α \tan 2 α \tan α}$ $1 = \frac{1}{\tan 2 α \tan α} - \frac{1}{\tan 3 α \tan α} - \frac{1}{\tan 3 α \tan 2 α}$ $α = \arctan t$ $1 = \frac{1}{\frac{2 t}{1 - t^{2}} \times t} - \frac{1}{\frac{3 t - t^{3}}{3 t^{2} - 1} \times t} - \frac{1}{\frac{2 t}{1 - t^{2}} \times \frac{3 t - t^{3}}{3 t^{2} - 1}}$ $1 = \frac{1 - t^{2}}{2 t^{2}} - \frac{3 t^{2} - 1}{t^{2} (t^{2} - 3)} - \frac{(1 - t^{2}) (3 t^{2} - 1)}{2 t^{2} (t^{2} - 3)}$ $1 = \frac{(1 - t^{2}) (t^{2} - 3) - 2 (3 t^{2} - 1) - (1 - t^{2}) (3 t^{2} - 1)}{2 t^{2} (t^{2} - 3)}$ $1 = \frac{(1 - t^{2}) (t^{2} - 3) - (3 t^{2} - 1) (2 + 1 - t^{2})}{2 t^{2} (t^{2} - 3)}$ $1 = \frac{(1 - t^{2}) (t^{2} - 3) - (3 t^{2} - 1) (3 - t^{2})}{2 t^{2} (t^{2} - 3)}$ $1 = \frac{(t^{2} - 3) ((1 - t^{2}) + (3 t^{2} - 1))}{2 t^{2} (t^{2} - 3)}$ $1 = \frac{(t^{2} - 3) 2 t^{2}}{2 t^{2} (t^{2} - 3)}$ $1 = 1 proved$
	Commented by kunal1234523 last updated on 27/Jun/18

	$r e a l l y s i m p l e a n d c o o l . t h e r e y o u h a d b e e n$ $t a k e n t = t a n α, i t h i n k$
	Commented by MJS last updated on 27/Jun/18

	$yes .$
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