N ninatemple New member Joined Jun 25, 2015 Messages 8 Jun 25, 2015 #1 1. 1/sin x - sin x = cos x/tan x2. 1/cos + tan x = cos x/1-sin x 3. tan x sin x/ 1-cos x = 1 + 1/cos x
1. 1/sin x - sin x = cos x/tan x2. 1/cos + tan x = cos x/1-sin x 3. tan x sin x/ 1-cos x = 1 + 1/cos x
D Deleted member 4993 Guest Jun 25, 2015 #2 ninatemple said: 1. 1/cos x + tan x = cos x/[1-sinx] ... use those [] to indicate order of operations .... very very important!!2. tan x sin x/[1-cos x] = 1 + 1/cos x 3. 1/sin x - sin x = cos x/tan x ...... see http://www.freemathhelp.com/forum/threads/91857-Need-Help-With-Proving-Trig-Identities Click to expand... 1/cos x + tan x = cos x/[1-sinx] L.H.S. = 1/cos(x) + tan(x) = 1/cos(x) + sin(x)/cos(x) = [1 + sin(x)]/cos(x) = [1 + sin(x)]/cos(x) * [1 - sin(x)]/[1 - sin(x)] = [1 - sin2(x)]/{cos(x)*[1 - sin(x)]} .... continue.... 2. tan x sin x/[1-cos x] = 1 + 1/cos x L.H.S. = tan x * sin x/[1-cos x] = tan x * sin x/[1-cos x] * [1+ cos x]/[1+ cos x] = tan x * sin x * [1+ cos x]/[1- cos2x] ....... continue...... Last edited by a moderator: Jun 25, 2015
ninatemple said: 1. 1/cos x + tan x = cos x/[1-sinx] ... use those [] to indicate order of operations .... very very important!!2. tan x sin x/[1-cos x] = 1 + 1/cos x 3. 1/sin x - sin x = cos x/tan x ...... see http://www.freemathhelp.com/forum/threads/91857-Need-Help-With-Proving-Trig-Identities Click to expand... 1/cos x + tan x = cos x/[1-sinx] L.H.S. = 1/cos(x) + tan(x) = 1/cos(x) + sin(x)/cos(x) = [1 + sin(x)]/cos(x) = [1 + sin(x)]/cos(x) * [1 - sin(x)]/[1 - sin(x)] = [1 - sin2(x)]/{cos(x)*[1 - sin(x)]} .... continue.... 2. tan x sin x/[1-cos x] = 1 + 1/cos x L.H.S. = tan x * sin x/[1-cos x] = tan x * sin x/[1-cos x] * [1+ cos x]/[1+ cos x] = tan x * sin x * [1+ cos x]/[1- cos2x] ....... continue......