Trigonometry & Inverse Trigonometry
Trigonometric Ratios and Identities
nta_pyq_2025_apr
Grade 12
Question:
If $\dfrac{\pi}{2} \leq x \leq \dfrac{3\pi}{4}$, then $\cos^{-1}\!\left(\dfrac{12}{13}\cos x + \dfrac{5}{13}\sin x\right)$ is equal to
$x - \tan^{-1}\!\dfrac{4}{3}$
$x + \tan^{-1}\!\dfrac{4}{5}$
$x - \tan^{-1}\!\dfrac{5}{12}$
$x + \tan^{-1}\!\dfrac{5}{12}$
Step-by-Step Solution
Key Concept: Write $\tfrac{12}{13}=\cos\alpha$ and $\tfrac{5}{13}=\sin\alpha$ where $\alpha=\tan^{-1}(5/12)$; the expression becomes $\cos^{-1}[\cos(x-\alpha)]$; verify $x-\alpha\in[0,\pi]$ on the given domain so $\cos^{-1}[\cos(x-\alpha)]=x-\alpha$.
Let $\alpha=\tan^{-1}(5/12) \in (0,\pi/2)$, so $\cos\alpha=\tfrac{12}{13}$, $\sin\alpha=\tfrac{5}{13}$. Then $\tfrac{12}{13}\cos x+\tfrac{5}{13}\sin x = \cos(x-\alpha)$. For $x\in[\pi/2,3\pi/4]$, $x-\alpha\in(0,\pi)$, so $\cos^{-1}[\cos(x-\alpha)]=x-\alpha=x-\tan^{-1}\!\left(\tfrac{5}{12}\right)$.
Correct Answer: 3