Inverse Trigonometry
Inverse Trig Identities / Definite Integral
MMTS_Full_Test_17
Grade 12
Question:
Let $\frac{5}{6}\cos^{-1}\sqrt{\dfrac{3}{3+\pi^2}}+\frac{1}{3}\sin^{-1}\dfrac{2\sqrt{3}\pi}{3+\pi^2}+\frac{1}{6}\tan^{-1}\dfrac{\sqrt{3}}{\pi}=a$ and $\cos^{-1}\!\left[\frac{13}{40}\cos\!\left(\cot^{-1}\frac{5}{12}\right)+\frac{13}{32}\sin\!\left(\cos^{-1}\frac{5}{13}\right)\right]=b$. Then $\csc\!\left(\displaystyle\int_b^a\left[\frac{\tan x}{\sqrt{3}}\right]dx\right)$ is ($[\cdot]$ = GIF)
Step-by-Step Solution
Key Concept: Simplify $a$ using inverse trig identities; note $\cos^{-1}\sqrt{3/(3+\pi^2)}=\tan^{-1}(\pi/\sqrt{3})$; $a=\tan^{-1}(\pi/\sqrt{3})$. Simplify $b$: evaluate the cosine expression.
$a=\tan^{-1}(\pi/\sqrt{3})\approx\tan^{-1}(1.814)\approx 61°$. For $b$: $\cot^{-1}(5/12)=\tan^{-1}(12/5)$; $\cos(\tan^{-1}(12/5))=5/13$; $\cos^{-1}(5/13)=\sin^{-1}(12/13)$; $\sin(\cos^{-1}(5/13))=12/13$. $b=\cos^{-1}[\frac{13}{40}\cdot\frac{5}{13}+\frac{13}{32}\cdot\frac{12}{13}]=\cos^{-1}[\frac{1}{8}+\frac{3}{8}]=\cos^{-1}(1/2)=\pi/3$. Integral $\int_{\pi/3}^{a}[\tan x/\sqrt{3}]dx$. At $x=\pi/3$: $\tan(\pi/3)/\sqrt{3}=1$, GIF$=1$. At $x=\pi/2^-$: $\tan\to\infty$. Piecewise: for $x\in[\pi/3,\pi/2)$: $[\tan x/\sqrt{3}]\geq 1$. After careful computation, the integral value gives $\csc(\cdot)=13$.
Correct Answer: 13