Answer: B
Given: \( \cos\theta = p \) ⇒ \( \sec\theta = \frac{1}{p} \) ⇒ \( \sec^2\theta = \frac{1}{p^2} \) Also, \( \cot\theta = q \) ⇒ \( \tan\theta = \frac{1}{q} \) ⇒ \( \tan^2\theta = \frac{1}{q^2} \) We know: \( \sec^2\theta - \tan^2\theta = 1 \) ⇒ \( \frac{1}{p^2} - \frac{1}{q^2} = 1 \)
Given: \( \cos\theta = p \) ⇒ \( \sec\theta = \frac{1}{p} \) ⇒ \( \sec^2\theta = \frac{1}{p^2} \) Also, \( \cot\theta = q \) ⇒ \( \tan\theta = \frac{1}{q} \) ⇒ \( \tan^2\theta = \frac{1}{q^2} \) We know: \( \sec^2\theta - \tan^2\theta = 1 \) ⇒ \( \frac{1}{p^2} - \frac{1}{q^2} = 1 \)