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# Choose the correct option. Justify your choice : (i) $$9sec^2A - 9tan^2A \ =$$ (A) 1 (B) 9 (C) 8 (D) 0 (ii) $$(1+ tan\theta + sec\theta )(1 + cos \theta – cosec \theta) \ =$$ (A) 0 (B) 1 (C) 2 (D) None of these (iii) $$(secA + tanA)(1 – sinA) \ =$$ (A) secA (B) sinA (C) cosecA (D) cosA (iv) $$\frac{1+tan^2A}{1+cot^2A} \ =$$ (A) $$sec^2A$$ (B) –1 (C) $$cot^2A$$ (D) none of these

(i) (B), as

$$9sec^2A - 9tan^2A$$
= $$9(sec^2A - tan^2A)$$
= $$9 × 1 \ = \ 9$$
[ $$\because 1+tan^2A \ = \ sec^2A$$ ]

(ii) (C), as

$$(1+ tan\theta + sec\theta )(1 + cos \theta – cosec \theta)$$

$$= \ (1 \ + \ \frac{sin\theta}{cos\theta} \ + \ \frac{1}{cos\theta} )(1 \ + \ \frac{cos\theta}{sin\theta} \ - \ \frac{1}{sin\theta})$$

$$= \ ( \frac{cos\theta \ + \ sin\theta \ + \ 1}{cos\theta})( \frac{sin\theta \ + \ cos\theta \ - \ 1}{sin\theta})$$

$$= \ \frac{(cos\theta + sin\theta )^2 \ - \ 1}{sin\theta cos\theta}$$
[$$\because (A+B)(A-B) \ = \ A^2 \ - \ B^2$$ ]

$$= \ \frac{cos^2 \theta \ + \ sin^2 \theta \ + \ 2cos\theta sin\theta \ - \ 1}{sin\theta cos\theta}$$
[ $$\because sin^2 \theta \ + \ cos^2 \theta \ = \ 1$$ ]

$$= \ \frac{1 \ + \ 2cos \theta sin \theta \ - \ 1}{sin \theta cos \theta} \ = \ \frac{2cos \theta sin\theta}{sin \theta cos \theta} \ = \ 2$$

(iii) (D), as

$$(secA + tanA) (1 – sinA) \$$
$$= \ ( \frac{1}{cosA} \ + \ \frac{sinA}{cosA})(1-sinA)$$

$$= \ ( \frac{1+sinA}{cosA})(1-sinA)$$
[ $$\because (A+B)(A-B) \ = \ A^2 \ - \ B^2$$ ]

$$= \ \frac{1 \ - \ sin^2A}{cosA} \ = \ \frac{cos^2A}{cosA} \ = \ cosA$$
[ $$\because sin^2A \ + \ cos^2A \ = \ 1$$ ]

(iv) (D), as

$$\frac{1+tan^2A}{1+cot^2A}$$ $$= \ \frac{sec^2A}{cosec^2A}$$
[ $$\because 1+tan^2A \ = \ sec^2A$$ and $$1+cot^2A \ = \ cosec^2A$$ ]

$$= \ \frac{ \frac{1}{cos^2A}}{ \frac{1}{sinn^2A}}$$ $$= \ \frac{sin^2A}{cos^2A} \ = \ tan^2A$$