Prove the identity $$ ext{cot}^2 heta - ext{cos}^2 heta = ext{cot}^2 heta ext{cos}^2 heta$$ - AQA - A-Level Maths Mechanics - Question 13 - 2017 - Paper 1

Substituting into LHS gives:

$\frac{ ext{cos}^2 heta}{ ext{sin}^2 heta} - ext{cos}^2 heta$

To combine these, express the second term with a common denominator:

$= \frac{ ext{cos}^2 heta}{ ext{sin}^2 heta} - \frac{ ext{cos}^2 heta ext{sin}^2 heta}{ ext{sin}^2 heta}$

This simplifies to:

$= \frac{ ext{cos}^2 heta - ext{cos}^2 heta ext{sin}^2 heta}{ ext{sin}^2 heta}$.

Factor out ext{cos}^2 heta:

$= \frac{ ext{cos}^2 heta(1 - ext{sin}^2 heta)}{ ext{sin}^2 heta}$

Using the identity $1 - ext{sin}^2 heta = ext{cos}^2 heta$, we get:

$= \frac{ ext{cos}^2 heta ext{cos}^2 heta}{ ext{sin}^2 heta}$

Which is the same as:

$= ext{cot}^2 heta ext{cos}^2 heta$

Thus, $ext{cot}^2 heta - ext{cos}^2 heta = ext{cot}^2 heta ext{cos}^2 heta$

This completes the proof.