Question

Prove that:

tan² θ + cot² θ = sec² θ cosec² θ − 2

Answer 1

In the given question, we need to prove tan² θ + cot² θ = sec² θ cosec² θ − 2

Now using `tan θ = sin θ/cos θ` and `cot θ = cos θ/sin θ` in LHS, we get

`tan^2 θ + cot^2  θ = sin^2 θ/cos^2 θ + cos^2 θ/sin^2 θ`

`= (sin^4 θ + cos^4 θ)/(cos^2 θ sin^2 θ)`

`= ((sin^2 θ)^2 + (cos^2 θ)^2)/(cos^2 θ sin^2 θ)`

Further, using the identity `a^2 + b^2 = (a + b)^2 - 2ab` we get

`((sin^2 θ)^2 + (cos^2 θ)^2)/(cos^2 θ sin^2 θ) = ((sin^2 θ + cos^ θ)^2 - 2 sin^2 θ cos^2 θ)/(sin^2 θ cos^2 θ)`

`= ((1)^2 - 2sin^2 θ cos^2 θ)/(sin^2 θ cos^2 θ)`

`= 1/(sin^2 θ cos^2 θ) - (2 sin^2 θ cos^2 θ)/(sin^2 θ cos^2 θ`

`= cosec^2 θ sec^2 θ - 2`

Since L.H.S = R.H.S

Hence proved.