Question

Prove that: `(sec^3 θ)/(sec^2 θ - 1) + ("cosec"^3 θ)/("cosec"^2 θ - 1) = sec θ . "cosec"  θ (sec θ + "cosec"  θ)`

Answer 1

Given to prove:

`(sec^3 θ)/(sec^2 θ - 1) + ("cosec"^3 θ)/("cosec"^2 θ - 1) = sec θ . "cosec"  θ (sec θ + "cosec"  θ)`

Recall the identities:

sec²θ – 1 = tan²θ

cosec²θ – 1 = cot²θ

Rewrite the left-hand side (LHS) using these:

`(sec^3θ)/(tan^2θ) + ("cosec"^3θ)/(cot^2θ)`

Recall:

`sec θ = 1/cos θ`

`"cosec"  θ = 1/sin θ`

`tan θ = sin θ/cos θ`

`cot θ = cos θ/sin θ`

Substitute these into the expression:

`(1/(cos θ))^3/((sin θ)/(cos θ))^2 + (1/(sin θ))^3/((cos θ)/(sin θ))^2`

Simplify each term:

First term:

`(1/(cos^3θ))/((sin^2θ)/(cos^2θ)) = 1/(cos^3θ) xx (cos^2θ)/(sin^2θ)`

= `1/(cos θ sin^2 θ)`

Second term:

`(1/(sin^3θ))/((cos^2θ)/(sin^2θ)) = 1/(sin^3θ) xx (sin^2θ)/(cos^2θ)`

= `1/(sin θ cos^2 θ)`

So the LHS becomes:

`1/(cos θ sin^2 θ) + 1/(sin θ cos^2 θ)`

The common denominator is sin² θ cos² θ.

Rewrite each term:

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

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

Actually, to get common denominator:

Multiply numerator and denominator appropriately:

First term:

`1/(cos θ sin^2 θ) = (cos θ)/(cos^2 θ sin^2 θ)`   ...(Multiply numerator and denominator by cos θ)

Second term:

`1/(sin θ cos^2 θ) = (sin θ)/(sin^2 θ cos^2 θ)`   ...(Multiply numerator and denominator by sin θ)

Now add:

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

RHS is:

`sec θ . "cosec"  θ . (sec θ + "cosec" θ) = 1/(cos θ) xx 1/(sin θ) xx (1/(cos θ) + 1/(sin θ))`

Simplify inside the bracket:

`1/(cos θ) + 1/(sin θ) = (sin θ + cos θ)/(sin θ cos θ)`

So RHS becomes:

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

Both sides are equal:

`(sin θ + cos θ)/(sin^2 θ cos^2 θ)`

Hence, the identity is proved.