Proving identities problem?

Question

(secx)(cscx) - tanx = (1-(tan^2)x)/(2(tanx)) + (1)/((2sinx)cosx))

I am helping my daughter study for exams and looked up questions online. I came across this and I can't seem to find the answers for this. We're really having a hard time answering this. Thank you so much for the help. :)

Answer 1

Here is another approach, perhaps a little easier

Write the both sides in terms of all sines and cosines (the rhs is messy but will eventually reduce to cot x and so will the lhs...just as Tom K. showed above.

Answer 2

The first thing I noticed was that the right side showed two trig. angle doubling formulas.

For the left side, (sec x)(csc x) - tan x = 1/((cos x)(sin x)) - (sin x)/(cos x) = (1 -sin^2(x))/(cosx sinx) = cos^2(x)/(cosx sinx) = cos x/sin x = cot x

The right hand side is (1 - tan^2 x)/(2 tan x) +1/(2 sin x cos x) = 1/(tan 2x) + 1/(sin 2x) = (cot 2x) + (csc 2x) = cos 2x/sin 2x + 1/sin 2x = (cos 2x + 1)/sin 2x = (2 cos^2 x - 1 + 1)/(2 sin x cos x) = 2 cos^2x/(2 sin x cos x) = cos x/sin x = cot x

SInce we have proven that both sides equal cot x, we can either say: done or we can reverse the steps of the right side.

We would then complete the left side with

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

cos 2x/sin 2x + 1/sin 2x = cot 2x + csc 2x = 1/tan 2x + 1/sin 2x = 1/(tan 2x/(1 - tan^2 2x) + 1/(2 sin x cos x) =

(1 - tan^2 2x)/(2 tan x) + 1/(2 sin x cos x)

Then, the complete proof, left to right, is

(sec x)(csc x) - tan x = 1/((cos x)(sin x)) - (sin x)/(cos x) = (1 -sin^2(x))/(cosx sinx) = cos^2(x)/(cosx sinx) = cos x/sin x = cot x = cos x/sin x = 2 cos^2 x/2 cos x sin x = (2 cos^2 x - 1 + 1)/(2 sin x cos x) = (cos 2x + 1)/sin 2x =

cos 2x/sin 2x + 1/sin 2x = cot 2x + csc 2x = 1/tan 2x + 1/sin 2x = 1/(tan 2x/(1 - tan^2 2x) + 1/(2 sin x cos x) =

(1 - tan^2 2x)/(2 tan x) + 1/(2 sin x cos x)