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Answer :
Given, \( sec\theta \ = \ \frac{13}{12} \ = \ \frac{H}{B} \ = \ \frac{AC}{AB} \)
Let AC = 13k and AB = 12k.
Then by Pythagoras theorem,
\( BC \ = \ \sqrt{AC^2 - AB^2} \ \)
\( = \ \sqrt{(13k)^2 - (12k)^2} \ \)
\( = \ \sqrt{169k^2 - 144k^2} \ \)
\(= \ \sqrt{25k^2} \ = \ 5k \)
\( \therefore sin\theta \ = \ \frac{P}{H} \ = \ \frac{BC}{AC} \ = \ \frac{5k}{13k} \ = \ \frac{5}{13} \)
\( cos\theta \ = \ \frac{B}{H} \ = \ \frac{AB}{AC} \ = \ \frac{12k}{13k} \ = \ \frac{12}{13} \)
\( tan\theta \ = \ \frac{P}{B} \ = \ \frac{BC}{AB} \ = \ \frac{5}{12} \)
\( cot\theta \ = \ \frac{1}{tan\theta} \ = \ \frac{12}{5} \)
\( cosec\theta \ = \ \frac{1}{sin\theta} \ = \ \frac{13}{5} \)