Proving identities

[KingAce]

problem: (tanx + cotx)(cosx + sinx) = secx +cscx

- I was able to simplify the equation to: (cscxsecx)(cosxsinx)=secx +cscx
and then, sin + (sin^2/cos) + (cos^2/sin) + cos = secx +cscx

what do I do next do prove the identity?

Also, a second problem, csc2x-cot2x = tanx is giving me a lot of trouble. How would I solve this one?

Thanks.

 

[morson]

LHS = (tanx + cotx)(cosx + sinx), RHS = secx + cscx

RHS = 1/cosx + 1/sinx = (sinx + cosx)/cosxsinx

LHS = tanxcosx + tanxsinx + cotxcosx + cotxsinx

= (sinx/cosx)cosx + (sinx/cosx)sinx + (cosx/sinx)cosx + (cosx/sinx)sinx

= sinx + (sinx)^2/cosx + (cosx)^2/sinx + cosx

= {cosx(sinx)^2 + (sinx)^3 + (cosx)^3 + sinx(cosx)^2}/sinxcosx

... after raising all to a common denominator

Grouping the numerator, the numerator becomes:

(sinx)^2[cosx + sinx] + (cosx)^2[cosx + sinx] = [(sinx)^2 + (cosx)^2][cosx + sinx]

= cosx + sinx, since [(sinx)^2 + (cosx)^2] = 1

So, now that the numerator is in simplest form, LHS = (sinx + cosx)/cosxsinx

= RHS

A true identity for an infinite set of values for x, but also not true for an infinite set of values for x (for example: x = 0).

 

[soroban]

Hello, KingAce!

For #2, we need some Double- and Half-Angle Identities:
. . \(\displaystyle \sin2x \:=\:2\cdot\sin x\cdot\cos x\)
. . \(\displaystyle \sin^2x \:=\:\frac{1\,-\,\cos2x}{2}\;\;\Rightarrow\;\;1\,-\,\cos2x \:=\:2\cdot\sin^2x\)

\(\displaystyle 2)\;\csc2x\,-\,\cot2x \:=\:\tan x\)

The left side is:
. . \(\displaystyle \L\:\frac{1}{\sin2x}\,-\,\frac{\cos2x}{\sin2x} \;=\;\frac{1\,-\,\cos2x}{\sin2x} \;=\;\frac{2\cdot\sin^2x}{2\cdot\sin x\cdot cos x} \;=\;\frac{\sin x}{\cos x} \;=\;\tan x\)