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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


Answer :


(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 \( 1+cot^2A \ = \ cosec^2A \) ]

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

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