Solveeit Logo
Login

Question

Question: If $\sin^{-1}(\sin\frac{7}{2}) + \cos^{-1}(\cos\frac{9}{2}) + \tan^{-1}(\tan\frac{11}{2}) + \cot^{-1...

If sin⁡−1(sin⁡72)+cos⁡−1(cos⁡92)+tan⁡−1(tan⁡112)+cot⁡−1(cot⁡132)=aπ+b\sin^{-1}(\sin\frac{7}{2}) + \cos^{-1}(\cos\frac{9}{2}) + \tan^{-1}(\tan\frac{11}{2}) + \cot^{-1}(\cot\frac{13}{2}) = a\pi + b, then (a+b)(a+b) equals

A

1

B

2

C

3

D

4

Answer

3

Explanation

Solution

Let E=sin⁡−1(sin⁡72)+cos⁡−1(cos⁡92)+tan⁡−1(tan⁡112)+cot⁡−1(cot⁡132)E = \sin^{-1}(\sin\frac{7}{2}) + \cos^{-1}(\cos\frac{9}{2}) + \tan^{-1}(\tan\frac{11}{2}) + \cot^{-1}(\cot\frac{13}{2}).

We need to evaluate each term using the properties of inverse trigonometric functions.

The relevant properties are:

sin⁡−1(sin⁡x)=x\sin^{-1}(\sin x) = x if x∈[−π2,π2]x \in [-\frac{\pi}{2}, \frac{\pi}{2}]

cos⁡−1(cos⁡x)=x\cos^{-1}(\cos x) = x if x∈[0,π]x \in [0, \pi]

tan⁡−1(tan⁡x)=x\tan^{-1}(\tan x) = x if x∈(−π2,π2)x \in (-\frac{\pi}{2}, \frac{\pi}{2})

cot⁡−1(cot⁡x)=x\cot^{-1}(\cot x) = x if x∈(0,π)x \in (0, \pi)

We use the approximation π≈3.1416\pi \approx 3.1416. π2≈1.5708\frac{\pi}{2} \approx 1.5708, π≈3.1416\pi \approx 3.1416, 3π2≈4.7124\frac{3\pi}{2} \approx 4.7124, 2π≈6.28322\pi \approx 6.2832.

Term 1: sin⁡−1(sin⁡72)\sin^{-1}(\sin\frac{7}{2})

72=3.5\frac{7}{2} = 3.5. The range of sin⁡−1(sin⁡x)\sin^{-1}(\sin x) is [−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}].

Since π2<72<3π2\frac{\pi}{2} < \frac{7}{2} < \frac{3\pi}{2} (i.e., 1.5708<3.5<4.71241.5708 < 3.5 < 4.7124), we use the formula sin⁡−1(sin⁡x)=π−x\sin^{-1}(\sin x) = \pi - x.

So, sin⁡−1(sin⁡72)=π−72\sin^{-1}(\sin\frac{7}{2}) = \pi - \frac{7}{2}.

Check if π−72∈[−π2,π2]\pi - \frac{7}{2} \in [-\frac{\pi}{2}, \frac{\pi}{2}].

π−72≈3.1416−3.5=−0.3584\pi - \frac{7}{2} \approx 3.1416 - 3.5 = -0.3584.

Since −1.5708≤−0.3584≤1.5708-1.5708 \le -0.3584 \le 1.5708, this is correct.

Term 2: cos⁡−1(cos⁡92)\cos^{-1}(\cos\frac{9}{2})

92=4.5\frac{9}{2} = 4.5. The range of cos⁡−1(cos⁡x)\cos^{-1}(\cos x) is [0,π][0, \pi].

Since π<92<2π\pi < \frac{9}{2} < 2\pi (i.e., 3.1416<4.5<6.28323.1416 < 4.5 < 6.2832), we use the formula cos⁡−1(cos⁡x)=2π−x\cos^{-1}(\cos x) = 2\pi - x.

So, cos⁡−1(cos⁡92)=2π−92\cos^{-1}(\cos\frac{9}{2}) = 2\pi - \frac{9}{2}.

Check if 2π−92∈[0,π]2\pi - \frac{9}{2} \in [0, \pi].

2π−92≈6.2832−4.5=1.78322\pi - \frac{9}{2} \approx 6.2832 - 4.5 = 1.7832.

Since 0≤1.7832≤3.14160 \le 1.7832 \le 3.1416, this is correct.

Term 3: tan⁡−1(tan⁡112)\tan^{-1}(\tan\frac{11}{2})

112=5.5\frac{11}{2} = 5.5. The range of tan⁡−1(tan⁡x)\tan^{-1}(\tan x) is (−π2,π2)(-\frac{\pi}{2}, \frac{\pi}{2}).

Since 3π2<112<5π2\frac{3\pi}{2} < \frac{11}{2} < \frac{5\pi}{2} (i.e., 4.7124<5.5<7.85404.7124 < 5.5 < 7.8540), we need to subtract 2π2\pi from 112\frac{11}{2}.

112−2π\frac{11}{2} - 2\pi.

Check if 112−2π∈(−π2,π2)\frac{11}{2} - 2\pi \in (-\frac{\pi}{2}, \frac{\pi}{2}).

112−2π≈5.5−6.2832=−0.7832\frac{11}{2} - 2\pi \approx 5.5 - 6.2832 = -0.7832.

Since −1.5708<−0.7832<1.5708-1.5708 < -0.7832 < 1.5708, this is correct.

So, tan⁡−1(tan⁡112)=112−2π\tan^{-1}(\tan\frac{11}{2}) = \frac{11}{2} - 2\pi.

Term 4: cot⁡−1(cot⁡132)\cot^{-1}(\cot\frac{13}{2})

132=6.5\frac{13}{2} = 6.5. The range of cot⁡−1(cot⁡x)\cot^{-1}(\cot x) is (0,π)(0, \pi).

Since 2π<132<3π2\pi < \frac{13}{2} < 3\pi (i.e., 6.2832<6.5<9.42486.2832 < 6.5 < 9.4248), we need to subtract 2π2\pi from 132\frac{13}{2}.

132−2π\frac{13}{2} - 2\pi.

Check if 132−2π∈(0,π)\frac{13}{2} - 2\pi \in (0, \pi).

132−2π≈6.5−6.2832=0.2168\frac{13}{2} - 2\pi \approx 6.5 - 6.2832 = 0.2168.

Since 0<0.2168<3.14160 < 0.2168 < 3.1416, this is correct.

So, cot⁡−1(cot⁡132)=132−2π\cot^{-1}(\cot\frac{13}{2}) = \frac{13}{2} - 2\pi.

Now, sum the terms:

E=(π−72)+(2π−92)+(112−2π)+(132−2π)E = (\pi - \frac{7}{2}) + (2\pi - \frac{9}{2}) + (\frac{11}{2} - 2\pi) + (\frac{13}{2} - 2\pi)

E=(π+2π−2π−2π)+(−72−92+112+132)E = (\pi + 2\pi - 2\pi - 2\pi) + (-\frac{7}{2} - \frac{9}{2} + \frac{11}{2} + \frac{13}{2})

E=(3π−4π)+(−7−9+11+132)E = (3\pi - 4\pi) + (\frac{-7 - 9 + 11 + 13}{2})

E=−π+(−16+242)E = -\pi + (\frac{-16 + 24}{2})

E=−π+82E = -\pi + \frac{8}{2}

E=−π+4E = -\pi + 4

E=4−πE = 4 - \pi

The given form is aπ+ba\pi + b.

Comparing 4−π4 - \pi with aπ+ba\pi + b, we get a=−1a = -1 and b=4b = 4.

We need to find (a+b)(a+b).

a+b=−1+4=3a+b = -1 + 4 = 3.