Question

If tan x, tan y, tan z are in G.P., show that cos 2y = cos(x+2)/cos(x -y)​

Answer 1

$\large{ \green { \rm GIVEN :-}}$

$\sf \tan(x) \: \: \: \tan(y) \: and \: \tan(z) \: are \: in \: G.P$

$\large{ \green { \rm \: TO \: \: SHOW :- }}$

$\sf \cos(2y) = \frac{ \cos( x + z) }{ \cos( x - z) }$

$\large{ \green { \rm SOLUTION:-}}$

$\sf \: If \: \: a,b,c \: \: are \: \: in \: \: G.P \: \: then : -$

$\sf \: relation \: is \\ \boxed{ \large\sf {b}^{2} = ac}$

$\sf \: Here \: \: a = \tan(x) \\ \: \: \: \: \: \: \: \: \: \: \: \: \sf b = \tan(y) \\ \: \: \: \: \: \: \: \: \: \: \: \sf \: c = \tan(z)$

$\sf \: Putting \: \: in \: \: the \: \: relation : -$

$\sf \: we \: get -$

$\: \: \: \: \: \: \: \: \sf \tan^{2} (y) = \tan(x) . \tan(y)$

$\sf \implies \frac{ \sin^{2} (y) }{ \cos ^{2} (y) } = \frac{ \sin(x) }{ \cos(x) } . \frac{ \sin(z) }{ \cos(z) }$

$\sf \: Appling \: \: Componando \: \: and \: \: Dividendo$

$\sf \implies \: \frac{ \sin^{2} (y) + \cos^{2}(y) }{ \sin^{2}(y) - \cos^{2}(y) } = \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \sf \: \frac{ \sin(x). \sin(z) + \cos(x) . \cos(z) }{ \sin(x). \sin(z) - \cos(x) . \cos(z)}$

$\sf \: Rearrange \: \: this$

$\sf \implies \: \frac{ \sin^{2} (y) + \cos^{2}(y) }{ - (\cos^{2} (y) - \sin^{2}(y) ) } = \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \sf \: \frac{ \cos(x) . \cos(z) + \sin(x). \sin(z) }{ - (\cos(x) . \cos(z) - \sin(x). \sin(z) ) }$

$\sf \: Removing \: or \: cancelling \: minus \: sign \:$

$\sf \implies \: \frac{ \sin^{2} (y) + \cos^{2}(y) }{ (\cos^{2} (y) - \sin^{2}(y) ) } = \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \sf \: \frac{ \cos(x) . \cos(z) + \sin(x). \sin(z) }{ (\cos(x) . \cos(z) - \sin(x). \sin(z) ) }.$

$\sf \: \bold{ Use \: \ Formula \rightarrow}$

$\sf \: 1.\sin^{2} (y) + \cos^{2}(y) = 1$

$2.\sf \: \cos(2y) = \cos^{2}(y) - \sin^{2} (y)$

$\sf \: 3. \cos(x - z) = \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \sf\cos(x) . \cos(z) + \sin(x). \sin(z)$

$4.\sf \: \cos(x + z) = \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \sf\cos(x) . \cos(z) - \sin(x). \sin(z)$

$\sf \: Putting \: \: all \: \: formula$

$\sf \: We \: \: get -$

$\implies \sf \: \frac{1}{ \cos(2y) } = \frac{ \cos(x - z) }{ \cos(x + z) }$

$\sf \: Now \: reciprocal \: this \: we \: get \: the \: ans$

$\boxed{ \sf \: \cos(2y) = \frac{ \cos(x + z) }{ \cos(x + z) } }$

$\large{ \green { \bold{ \sf \: Proved : - }}}$

Answer 2

Answer:

understand it by seeing