Question

help :)

trigonometry ....

proving identities....


please help...

Answer 1

\[\frac{ \tan x }{ 1+\cos x } + \frac{ \sin x }{ 1-\cos x } = \cot x +\sec x \csc x\]

Answer 2

@ash2326

Answer 3

@.Sam.

Answer 4

@ganeshie8

Answer 5

@agent0smith

Answer 6

I'd try combining the fractions on the left

Answer 7

yeah i also do that..

Answer 8

but still i dont get the right answer

Answer 9

\[\frac{ \tan \ x (1 - \cos x ) + \sin x ( 1+ \cos x) }{ (1 - \cos x)(1+ cosx) }\]\[\frac{ \tan ( 1- \cos) + \sin ( 1 + \cos )}{ 1 - \cos^{2} }\]\[\frac{ \tan ( 1 - \cos) + \sin (1+ \cos ) }{ \sin^2 }\]\[\frac{ \sin \ sec ( 1 - \cos ) + \sin (1 + \cos) }{ \sin^{2}}\]\[\frac{ sec ( 1- \cos) }{ \sin } + cosec ( 1 + \cos)\]\[\sec \ cosec \ - \sec \ \cot + cosec + cosec \ \cos\]
now simplify this... u will get ur answer ..
can u do that ?

Answer 10

=\[\frac{ \tan(1-cosx) + sinx(1+cosx) }{ (1+cosx) (1-cosx)}\]

=
\frac{ sinz }{ cosx }(1-cosx)+sinx(1+cosx)
-----------------------------------
1-cos^2x

=

Answer 11

\[\frac{ \tan \ x (1 - \cos x ) + \sin x ( 1+ \cos x) }{ (1 - \cos x)(1+ cosx) }\]\[\frac{ \tan ( 1- \cos) + \sin ( 1 + \cos )}{ 1 - \cos^{2} }\]\[\frac{ \tan ( 1 - \cos) + \sin (1+ \cos ) }{ \sin^2 }\]\[\frac{ \sin \ sec ( 1 - \cos ) + \sin (1 + \cos) }{ \sin^{2}}\]\[\frac{ sec ( 1- \cos) }{ \sin } + cosec ( 1 + \cos)\]\[\sec \ cosec \ - \sec \ \cot + cosec + cosec \ \cos\]

Answer 12

did u get ur answer ?

Answer 13

no did'nt

Answer 14

its confusing

Answer 15

sorry

Answer 16

is it confusing to u from the beginning or is it after a certain step ?