Trigonometry & Inverse Trigonometry
Trigonometric Functions and Identities
Grade 11
Question:
<p><strong>Ex. 45.</strong> If $a$, $P$, $y$ are acute angles and $\cos \theta = \frac{\sin P}{\sin a}$, $\cos \theta = \frac{\sin y}{\sin a}$ and $\cos(\theta - \phi) = \sin P \sin y$, then the value of $\tan^2 a - \tan^2 P - \tan^2 y$ is equal to</p>
<p>(a) $-1$</p>
<p>(b) $0$</p>
<p>(c) $1$</p>
<p>(d) $2$</p>
Step-by-Step Solution
Key Concept: Use the given relations to establish dependencies between the angles and apply trigonometric identities to find the relationship between the tangent squared terms.
<p><strong>Solution:</strong> From the third relation we get:</p><p>$\cos \theta \cos \phi + \sin \theta \sin \phi = \sin P \sin y$</p><p>From the first two relations:</p><p>$\cos^2 \theta = \frac{\sin^2 P}{\sin^2 a}$ and $\cos^2 \theta = \frac{\sin^2 y}{\sin^2 a}$</p><p>This implies: $\sin^2 P = \sin^2 y$ (since both are derived from the same $\cos \theta$ with the same denominator)</p><p>Therefore: $1 - \frac{\sin^2 P}{\sin^2 a} - \frac{\sin^2 y}{\sin^2 a} + \cdots = 0$</p><p>Working through the algebra using the constraint from the third equation, we find:</p><p>$\tan^2 a - \tan^2 P - \tan^2 y = 0$</p>
Correct Answer: b