need some help with this trigo identities?

Question

Prove that (1 + tan x - sec x)( 1 + cot x + cosec x) = 2
thanks!

Answer 1

(1 + tan x - sec x)( 1 + cot x + cosec x)
Multiply through:
1 + cot x + csc x + tan x + (tan x cot x) + (tan x cscx)
- sec x - (sec x cot x) - (sec x csc x)

Note:
tan x cot x = (sin / cos)(cos / sin) = 1
tan x csc x = (sin / cos)(1 / sin) = sec x
sec x cot x = (1 / cos)(cos / sin) = csc x

So, we have:
1 + cot x + csc x + tan x + 1 + sec x
- sec x - (csc x) - (sec x csc x)
= 2 + cot x + tan x - sec x csc x
= 2 + (cos / sin) + (sin / cos) - (1 / sin*cos)
= 2 + [(cos^2 + sin^2 - 1) / (sin*cos)]
= 2 + [ (1 - 1) / sin*cos]
= 2 + 0
= 2

Answer 2

(1 + tan x - sec x)( 1 + cot x + cosec x)
= (1 + sinx / cosx - 1/cosx) x (1 + cosx/sinx + 1/sinx)
= ( cosx + sinx - 1) x (sinx + cosx + 1) / sinx.cosx

= (cosxsinx + cos^2x + cosx + sin^2x + sinxcosx + sinx - sinx - cosx - 1) / sinxcosx

= (cos^2x + sin^2x + 2sinxcosx - 1) / sinxcosx

= ( 1 + 2sinxcosx - 1) / sinxcosx

= 2sinxcosx/sinxcosx

= 2

Proved

Answer 3

1. Move everything to sin and cos, using
tan = sin/cos
sec = 1/cos
cot = cos/sin
cosec = 1/sin

(1 + sin/cos - 1/cos)(1 + cos/sin + 1/sin)

2. Within each bracket, use a common denominator

(cos/cos + sin/cos - 1/cos)(sin/sin + cos/sin + 1/sin)

[(cos + sin - 1)/cos][(sin+cos+1)/sin]

3. Multiply:
(SinCos + Cos^2 +Cos +Sin^2 +SInCos + SIn - Sin -Cos -1)/SinCos

4. Simplify, using
Sin^2 + Cos^2 = 1

(2SinCos +1 -1)/SinCos =
2SinCos/SinCos =
2

Answer 4

sorry if this is hard to follow, but multiply everything out:
1+cot+csc+tan+tancot+tancsc-sec-seccot-seccsc
1+cos/sin+1/sin+sin/cos+1+1/cos-1/cos-1/sin-1/sincos
2+cos/sin+sin/cos-1/sincos
(2sincos+cos^2+sin^2-1)/sincos
(2sincos+(cos^2 +sin^2) -1)/sin cos
(2sincos+1-1)/sincos
2sincos/sincos
=2

Answer 5

(1 + (sin/cos) - (1/cos) ) ( 1 + (cos/sin) + (1/sin) )
= ((cos + sin - 1)/cos) ( (cos + sin +1) / sin)
= ((cos + sin)^2 - 1) / ( cos sin) ,
{ since numerator is of the form (a+b)(a-b), which equals a^2 - b^2}

= (cos^2 + 2 sin cos + sin^2 - 1) / (cos sin)
= (2 sin cos + 1 - 1) / (cos sin)
= (2 sin cos) / (cos sin)
= 2

Answer 6

http://www.projectalevel.co.uk/maths/seccoseccot.htm

Answer 7

(1+sinx/cosx-1/cosx)(1+cosx/sinx+1/sinx)
(cosx+sinx-1)(sinx+cosx+1)/sinx.cosx
[(sinx+cosx)^2 -1]/sinx.cosx
(sinx^2+cosx^2+2sinx.cosx-1)/sinx.cosx
(1+2sinx.cosx-1)/sinx.cosx
2sinx.cosx/sinx.cosx
2