Verified answer

secθ = 1/cosθ

LHS

= (secθ+1)/(secθ-1) + (cosθ+1)/(cosθ-1)

= (1/cosθ + 1)/(1/cosθ - 1) + (cosθ+1)/(cosθ-1)

= (1+cosθ)/(1-cosθ) + (cosθ+1)/(cosθ-1)

= (1+cosθ)/(1-cosθ) - (1+cosθ)/(1-cosθ)

= 0

= RHS

I think secθ +1 is over secθ - 1. the same with RHS it is cosθ +1 over cosθ -1.

( secθ +1) / (secθ - 1) + (cosθ +1)/(cosθ -1) =0

(1/cosθ +1)/(1/cosθ -1) + (cosθ +1)/(cosθ -1) =0

(1/cosθ +cosθ/cosθ)/(1/cosθ -cosθ/cosθ) + (cosθ +1)/(cosθ -1) =0

((1+cosθ)/cosθ)/((1-cosθ)/cosθ) + (cosθ +1)/(cosθ -1) =0

((1+cosθ)/((1-cosθ)) + (cosθ +1)/(cosθ -1) =0

In the 2nd term, If you reverse the sign of (cosθ -1), the sign before the term would become -

(1+cosθ)/((1-cosθ) - (cosθ +1)/(1 - cosθ) =0

0 = 0

LHS = (sec θ + 1 / sec θ - 1) + (cosθ + 1 / cosθ - 1)

now we know that secθ = 1/cos θ

replace that throughout and we get

= (1/cosθ + 1) / (1/cosθ - 1) + (cosθ + 1)/(cosθ-1)

= (1+cosθ)/(1-cosθ)/cosθ/cosθ) + (cosθ + 1)/(cosθ-1)

= - (1+cosθ)/(cosθ-1) + (1+cosθ)/(cosθ-1)

= 0

Hence proved

(I assume you mean...)

(secx + 1) / (secx - 1) + (cosx + 1) / (cosx - 1) = 0

(secx + 1) / (secx - 1) = - (cosx + 1) / (cosx - 1)

So we start at the LHS and try to get to the RHS

(secx + 1) / (secx - 1)

= ((1 / cosx) + 1) / ((1 / cosx) - 1)

= [(1 + cosx) / cosx] / [(1 - cosx) / cosx]

= (1 + cosx) / (1 - cosx)

= - (1 + cosx) / (cosx - 1)

as required

Twiddley-two, divide by 2TTr, carry the y, add the sum of tuppence cubed, therefore, NOT OUT!

sec0=1/COS0

(sec0+1/sec0-1)+(cos0+1/cos0-1)

{(1/cos0)+1)/((1/cos0-1)-1)}+(cos0+1/cos0-1)

((1+cos0/cos0)/cos0}/(1-cos0)/cos0

{(1+cos0)/1-cos0}+[1+cos0]/cos0-1

inorder to make the denominator same,

take - common outside in cos0-1

=-(1-cos0)

{(1+cos0)/1-cos0}+[[1+cos0]/-(1-cos0)]

{(1+cos0)/1-cos0}-[1+cos0]/(1-cos0)]

=0.

hence,the proof.

note that sec (theta) = 1/ cos (theta)

(sec (theta)+1 )/(sec (theta) -1) = (1+cos(theta))/(1-cos(theta))

= - (1+cos(theta))/(cos(theta)-1)

(sec(theta)+1)/(sec(theta)-1)+ (1+cos(theta))/(cos(theta)-1)=0