Prove that :

(cotx - tanx)/(cotx + tanx) = cosx^2 - sinx^2

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The identity `(cotx - tanx)/(cotx + tanx) = cos^2x - sin^2x` has to be proved.

Start with the left hand side

`(cotx - tanx)/(cotx + tanx)`

=> `((cosx/sinx) - (sinx/cosx))/((cosx/sinx)+(sinx/cosx))`

=> `((cos^2x - sin^2x)/(sinx*cosx))/((cos^2x + sin^2x)/(sinx*cosx))`

=> `(cos^2x - sin^2x)/1`

=> `cos^2x - sin^2x`

which is the right hand side

**This proves that `(cotx - tanx)/(cotx + tanx) = cos^2x - sin^2x` **

since cotx = cosx/sinx and tanx = sinx/cosx

so (cotx - tanx)/(cotx + tanx)

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

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

sinxcosx sinxcosx

= (cosx^2 - sinx^2) X sinxcosx

sinxcosx cosx^2 + sinx^2

sinxcosx should cancel each other

and as we know cosx^2 + sinx^2 = 1

so we have

(cosx^2 - sinx^2) X 1

1 1

= cosx^2 - sinx^2

Hence (cotx - tanx)/(cotx + tanx) = cosx^2 - sinx^2 proved

since cotx = cosx/sinx and tanx = sinx/cosx

so (cotx - tanx)/(cotx + tanx)

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

= [(cosx^2 - sinx^2)/sinxcosx] / [(cosx^2 + sinx^2)/sinxcosx]

= (cosx^2 - sinx^2)/sinxcosx X sinxcosx /(cosx^2 + sinx^2)

**sinxcosx should cancel each other**

and as we know cosx^2 + sinx^2 = 1

so we have

(cosx^2 - sinx^2)/1 X 1/1

= cosx^2 - sinx^2Hence (cotx - tanx)/(cotx + tanx) = cosx^2 - sinx^2 proved

L:H:S ≡ (cotx - tanx) ÷ (cotx + tanx)

= [(cos²x-sin²x)/sinx.cosx] ÷ [(cos²x+sin²x)/sinx.cosx]

= (cos²x-sin²x) ÷ (cos²x+sin²x)

**we know that sin²θ + cos²θ = 1**

= cos²x-sin²x

= R:H:S

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