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Q5. Given \( sec\theta \ = \ \frac{13}{12} \) , calculate all other trigonometric ratios.
Answer :


Given, \( sec\theta \ = \ \frac{13}{12} \ = \ \frac{H}{B} \ = \ \frac{AC}{AB} \)

Let AC = 13k and AB = 12k.

Then by Pythagoras theorem,

\( BC \ = \ \sqrt{AC^2 - AB^2} \ = \ \sqrt{(13k)^2 - (12k)^2} \ = \ \sqrt{169k^2 - 144k^2} \ = \ \sqrt{25k^2} \ = \ 5k \)



∴ \( sin\theta \ = \ \frac{P}{H} \ = \ \frac{BC}{AC} \ = \ \frac{5k}{13k} \ = \ \frac{5}{13} \)

\( cos\theta \ = \ \frac{B}{H} \ = \ \frac{AB}{AC} \ = \ \frac{12k}{13k} \ = \ \frac{12}{13} \)

\( tan\theta \ = \ \frac{P}{B} \ = \ \frac{BC}{AB} \ = \ \frac{5}{12} \)

\( cot\theta \ = \ \frac{1}{tan\theta} \ = \ \frac{12}{5} \)

\( cosec\theta \ = \ \frac{1}{sin\theta} \ = \ \frac{13}{5} \)