Inverse Trigonometry
tan of difference of arcsin and 2·arccos
MJAT_TS7_P2
Grade 12
Question:
Considering only principal values of inverse trigonometric functions, the value of $\tan\!\left(\sin^{-1}\frac{3}{5}-2\cos^{-1}\frac{2}{5}\right)$ is:
A) $\dfrac{7}{24}$
B) $-\dfrac{7}{24}$
C) $-\dfrac{5}{24}$
D) $\dfrac{5}{24}$
Step-by-Step Solution
Key Concept: Let $\alpha=\sin^{-1}(3/5)$: $\sin\alpha=3/5$, $\cos\alpha=4/5$, $\tan\alpha=3/4$. Let $\phi=\cos^{-1}(2/5)$: $\tan\phi=\sqrt{21}/2$. Find $\tan(2\phi)=2\tan\phi/(1-\tan^2\phi)=\sqrt{21}/(-17/4)=-4\sqrt{21}/17$. Then $\tan(\alpha-2\phi)$ using the subtraction formula.
Answer: **B** $\left(-\dfrac{7}{24}\right)$.
Correct Answer: B