Given that \(\displaystyle x \cos \theta = a \quad \text{and} \quad y = a \tan \theta \), then find the value of \(\displaystyle x^2 – y^2 \).
Solution:
From \( \displaystyle x \cos \theta = a \), we get: \(\displaystyle x = \frac{a}{\cos \theta} \)
and
\( \displaystyle y = a \tan \theta = a \cdot \frac{\sin \theta}{\cos \theta} \)
Therefore,
\( \begin{aligned} x^2 – y^2 = &\frac{a^2}{\cos^2 \theta} \,- a^2 \cdot \frac{\sin^2 \theta}{\cos^2 \theta} \\ &= \frac{a^2 \,- a^2 \sin^2 \theta}{\cos^2 \theta} \\ &= \frac{a^2 (1 \,- \sin^2 \theta)}{\cos^2 \theta} \\ &= \frac{a^2 \cos^2 \theta}{\cos^2 \theta}, \quad \because 1 \,- \sin^2 \theta = \cos^2 \theta \\ &= a^2 \end{aligned} \)The final answer is:
\( \boxed{x^2 – y^2 = a^2} \)Please let me know in the comments if you find any errors in this solution.