If \( x=\sec\theta+\cos\theta,\ y=\sec^{n}\theta+\cos^{n}\theta \), then \( \displaystyle\frac{dy}{dx} \) is equal to
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( n\left(\sqrt{\displaystyle\frac{y^2-4}{x^2-4}}\right) \)
Since x and y are both given in terms of θ, differentiate each with respect to θ.
\( \displaystyle \frac{dx}{d\theta}=\sec\theta\tan\theta-\sin\theta \)
\( \displaystyle =\tan\theta(\sec\theta-\cos\theta) \)
\( \displaystyle \frac{dy}{d\theta}=n\sec^{n-1}\theta\cdot\sec\theta\tan\theta+n\cos^{n-1}\theta\cdot(-\sin\theta) \)
\( \displaystyle =n\tan\theta(\sec^{n}\theta-\cos^{n}\theta) \)
\( \displaystyle \frac{dy}{dx}=\frac{n(\sec^{n}\theta-\cos^{n}\theta)}{\sec\theta-\cos\theta} \)
Now write the numerator and denominator in terms of y and x. Since \( \sec\theta\cos\theta=1 \):
\( \displaystyle (\sec^{n}\theta-\cos^{n}\theta)^2=(\sec^{n}\theta+\cos^{n}\theta)^2-4\sec^{n}\theta\cos^{n}\theta \)
\( \displaystyle =y^2-4 \)
\( \displaystyle (\sec\theta-\cos\theta)^2=(\sec\theta+\cos\theta)^2-4 \)
\( \displaystyle =x^2-4 \)
Hence
\( \displaystyle \frac{dy}{dx}=n\sqrt{\frac{y^2-4}{x^2-4}} \)
Why the other options are incorrect
Option (1) \( \displaystyle\frac{n^2(y^2+4)}{x^2+4} \) — Incorrect. The signs inside are wrong: \( (a-b)^2=(a+b)^2-4ab \) gives \( y^2-4 \) and \( x^2-4 \), and the square root is also missing.
Option (2) \( \displaystyle\frac{n^2(y^2-4)}{x^2-4} \) — Incorrect. This is \( \left(\frac{dy}{dx}\right)^2 \), not \( \frac{dy}{dx} \); the square root has not been taken.
Option (3) \( n\left(\sqrt{\displaystyle\frac{y^2+4}{x^2+4}}\right) \) — Incorrect. \( (a-b)^2=(a+b)^2-4ab \) gives \( y^2-4 \) and \( x^2-4 \), not \( y^2+4 \) and \( x^2+4 \).