Inverse Trigonometric FunctionsmediumFree

Inverse Trigonometric Functions: Lowest Form Find

JEE Maths question with a full step-by-step solution.

Question
If A=cot⁡−1(1)+12cot⁡−1(12)+13cot⁡−1(13)A=\cot^{-1}(1)+\dfrac{1}{2}\cot^{-1}\left(\dfrac{1}{2}\right)+\dfrac{1}{3}\cot^{-1}\left(\dfrac{1}{3}\right) and B=cot⁡−1(1)+2cot⁡−1(2)+3cot⁡−1(3)B=\cot^{-1}(1)+2\cot^{-1}(2)+3\cot^{-1}(3), then ∣B−A∣=aπb+cdcot⁡−1(3)|B-A|=\dfrac{a\pi}{b}+\dfrac{c}{d}\cot^{-1}(3) where a,b,c,d∈Na,b,c,d\in\mathbb{N} in lowest form. Find (a+b+c+d)(a+b+c+d).
Solution
Answer: 40
Step 1: Compute B−AB-A:
B−A=2cot⁡−12+3cot⁡−13−12cot⁡−112−13cot⁡−113B-A=2\cot^{-1}2+3\cot^{-1}3-\dfrac{1}{2}\cot^{-1}\dfrac{1}{2}-\dfrac{1}{3}\cot^{-1}\dfrac{1}{3}
 Step 2: Use cot⁡−11x=tan⁡−1x=π2−cot⁡−1x\cot^{-1}\dfrac{1}{x}=\tan^{-1}x=\dfrac{\pi}{2}-\cot^{-1}x for x>0x>0:
cot⁡−112=π2−cot⁡−12,cot⁡−113=π2−cot⁡−13\cot^{-1}\dfrac{1}{2}=\dfrac{\pi}{2}-\cot^{-1}2,\quad\cot^{-1}\dfrac{1}{3}=\dfrac{\pi}{2}-\cot^{-1}3
 Step 3: Substitute:
B−A=2cot⁡−12+3cot⁡−13−12(π2−cot⁡−12)−13(π2−cot⁡−13)B-A=2\cot^{-1}2+3\cot^{-1}3-\dfrac{1}{2}\left(\dfrac{\pi}{2}-\cot^{-1}2\right)-\dfrac{1}{3}\left(\dfrac{\pi}{2}-\cot^{-1}3\right)
=2cot⁡−12+3cot⁡−13−π4+12cot⁡−12−π6+13cot⁡−13=2\cot^{-1}2+3\cot^{-1}3-\dfrac{\pi}{4}+\dfrac{1}{2}\cot^{-1}2-\dfrac{\pi}{6}+\dfrac{1}{3}\cot^{-1}3
 Step 4: Collect like terms:
=(2+12)cot⁡−12+(3+13)cot⁡−13−π4−π6=\left(2+\dfrac{1}{2}\right)\cot^{-1}2+\left(3+\dfrac{1}{3}\right)\cot^{-1}3-\dfrac{\pi}{4}-\dfrac{\pi}{6}
=52cot⁡−12+103cot⁡−13−5π12=\dfrac{5}{2}\cot^{-1}2+\dfrac{10}{3}\cot^{-1}3-\dfrac{5\pi}{12}
 Step 5: Split: 103cot⁡−13=52cot⁡−13+(103−52)cot⁡−13=52cot⁡−13+56cot⁡−13\dfrac{10}{3}\cot^{-1}3=\dfrac{5}{2}\cot^{-1}3+\left(\dfrac{10}{3}-\dfrac{5}{2}\right)\cot^{-1}3=\dfrac{5}{2}\cot^{-1}3+\dfrac{5}{6}\cot^{-1}3. So:
B−A=52(cot⁡−12+cot⁡−13)+56cot⁡−13−5π12B-A=\dfrac{5}{2}(\cot^{-1}2+\cot^{-1}3)+\dfrac{5}{6}\cot^{-1}3-\dfrac{5\pi}{12}
 Step 6: Use cot⁡−12+cot⁡−13=π4\cot^{-1}2+\cot^{-1}3=\dfrac{\pi}{4}:
B−A=52⋅π4+56cot⁡−13−5π12=5π8−5π12+56cot⁡−13B-A=\dfrac{5}{2}\cdot\dfrac{\pi}{4}+\dfrac{5}{6}\cot^{-1}3-\dfrac{5\pi}{12}=\dfrac{5\pi}{8}-\dfrac{5\pi}{12}+\dfrac{5}{6}\cot^{-1}3
 Step 7: Combine the π\pi terms with common denominator 2424:
5π8−5π12=15π−10π24=5π24\dfrac{5\pi}{8}-\dfrac{5\pi}{12}=\dfrac{15\pi-10\pi}{24}=\dfrac{5\pi}{24}
 Step 8: Therefore:
∣B−A∣=5π24+56cot⁡−13=aπb+cdcot⁡−13|B-A|=\dfrac{5\pi}{24}+\dfrac{5}{6}\cot^{-1}3=\dfrac{a\pi}{b}+\dfrac{c}{d}\cot^{-1}3
 So a=5,b=24,c=5,d=6a=5,b=24,c=5,d=6. Step 9:
a+b+c+d=5+24+5+6=40a+b+c+d=5+24+5+6=40
 Answer: 4040
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