Prove that

$(i) (s i n^{2} A c o s^{2} B - c o s^{2} A s i n^{2} B) = (s i n^{2} A - s i n^{2} B)$

$(i i) (t a n^{2} A s e c^{2} B - s e c^{2} A t a n^{2} B) = (t a n^{2} A - t a n^{2} B)$ [2 MARKS]

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Solution

Concept: 0.5 Mark each
Application: 0.5 Mark each

We have

(i) LHS $= (s i n^{2} A c o s^{2} B - c o s^{2} A s i n^{2} B)$

$= s i n^{2} A (1 - s i n^{2} B) - (1 - s i n^{2} A) s i n^{2} B$

$= s i n^{2} A - s i n^{2} B =$ RHS

(ii) LHS $= (t a n^{2} A s e c^{2} B - s e c^{2} A t a n^{2} B)$

$= t a n^{2} A (1 + t a n^{2} B) - (1 + t a n^{2} A) t a n^{2} B$

$= (t a n^{2} A - t a n^{2} B) =$ RHS

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