Home › MCQs › EEJ MAIN Mathematics › Question #6967
Back to Questions
EEJ MAIN Mathematics QUESTION #6967
Question 1

Evaluate $\sin^{-1}\!\left(\dfrac{12}{13}\right)-\sin^{-1}\!\left(\dfrac{3}{5}\right)$.

  • $\pi-\sin^{-1}\!\left(\dfrac{63}{65}\right)$
  • $\dfrac{\pi}{2}-\sin^{-1}\!\left(\dfrac{56}{65}\right)$✔️
  • $\dfrac{\pi}{2}-\cos^{-1}\!\left(\dfrac{9}{65}\right)$
  • $\pi-\cos^{-1}\!\left(\dfrac{33}{65}\right)$
Correct Answer Explanation

Let $A=\sin^{-1}(12/13)$: $\sin A=12/13,\ \cos A=5/13$.

Let $B=\sin^{-1}(3/5)$: $\sin B=3/5,\ \cos B=4/5$.

$\sin(A-B)=\sin A\cos B-\cos A\sin B=\dfrac{12}{13}\cdot\dfrac{4}{5}-\dfrac{5}{13}\cdot\dfrac{3}{5}=\dfrac{48}{65}-\dfrac{15}{65}=\dfrac{33}{65}$... Hmm, but let me check $\cos(A-B)$: $=\cos A\cos B+\sin A\sin B=\dfrac{5}{13}\cdot\dfrac{4}{5}+\dfrac{12}{13}\cdot\dfrac{3}{5}=\dfrac{20+36}{65}=\dfrac{56}{65}$.

So $A-B=\cos^{-1}(56/65)=\pi/2-\sin^{-1}(56/65)$. Answer: $\dfrac{\pi}{2}-\sin^{-1}\!\left(\dfrac{56}{65}\right)$.