Question

Question #7dffd

Answer 1

Let us turn the left-hand side into the right-hand side of the equation.

`tan^2x-cot^2x`

by `tanx={sinx}/{cosx}` and `cotx={cosx}/{sinx}`,

`={sin^2x}/{cos^2x}-{cos^2x}/{sin^2x}`

by finding the common denominator,

`={sin^4x-cos^4x}/{cos^2x sin^2x}`

by factoring out the numerator,

`={(sin^2x-cos^2x)(sin^2x+cos^2x)}/{cos^2x sin^2x}`

by `sin^2x+cos^2x=1`,

`={sin^2x-cos^2x}/{cos^2x sin^2x`

by splitting into two fractions,

`={sin^2x}/{sin^2x cos^2x}-{cos^2x}/{sin^2x cos^2x}`

by cancelling out common factors,

`=1/{cos^2x}-1/{sin^2x}`

by `secx=1/cosx` and `cscx=1/sinx`,

`=sec^2x-csc^2x`

Hence,

`tan^2x-cot^2x=sec^2x-csc^2x`.

---

I hope that this was helpful.