Trigonometry & Inverse Trigonometry
tan(2sin⁻¹ − 2cos⁻¹) — Double Angle Identity
nta_pyq_2026_jan
Grade 12
Question:
Considering the principal values of inverse trigonometric functions, the value of $\tan\!\left(2\sin^{-1}\!\left(\dfrac{2}{\sqrt{13}}\right)-2\cos^{-1}\!\left(\dfrac{3}{\sqrt{10}}\right)\right)$ is equal to:
$\dfrac{16}{63}$
$-\dfrac{33}{56}$
$-\dfrac{16}{63}$
$\dfrac{33}{56}$
Step-by-Step Solution
Key Concept: Let $\alpha=\sin^{-1}\!\tfrac{2}{\sqrt{13}}$: $\sin\alpha=\tfrac{2}{\sqrt{13}}$, $\cos\alpha=\tfrac{3}{\sqrt{13}}$, $\tan\alpha=\tfrac{2}{3}$. $\tan2\alpha=\tfrac{12}{5}$. Let $\gamma=\cos^{-1}\!\tfrac{3}{\sqrt{10}}$: $\cos\gamma=\tfrac{3}{\sqrt{10}}$, $\tan\gamma=\tfrac{1}{3}$. $\tan2\gamma=\tfrac{3}{4}$.
$\tan(2\alpha-2\gamma)=\dfrac{33}{56}$.
Correct Answer: 4