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Inverse Trigonometric Functions question

2023 · 31 Jan · Shift 1 · Q27
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  5. /2023 · 31 Jan · Shift 1 · Q27

Inverse Trigonometric Functions question

2023 · 31 Jan · Shift 1 · Q27

JEE MainMathematicsInverse Trigonometric FunctionsMCQ+4 / −1
If sin⁡−1α17+cos⁡−145−tan⁡−17736=0,0<α<13{\sin ^{ - 1}}{\alpha \over {17}} + {\cos ^{ - 1}}{4 \over 5} - {\tan ^{ - 1}}{{77} \over {36}} = 0,0 \lt \alpha \lt 13sin−117α​+cos−154​−tan−13677​=0,0<α<13, then sin⁡−1(sin⁡α)+cos⁡−1(cos⁡α){\sin ^{ - 1}}(\sin \alpha ) + {\cos ^{ - 1}}(\cos \alpha )sin−1(sinα)+cos−1(cosα) is equal to :
  1. A
    16
  2. B
    π\piπ
  3. C
    16 −-− 5 π\piπ
  4. D
    0
View written solutionFree

Correct answer: B

  1. Given equation

We have

sin⁡−1(α17)+cos⁡−1(45)−tan⁡−1(7736)=0\sin^{-1}\left(\frac{\alpha}{17}\right)+\cos^{-1}\left(\frac45\right)-\tan^{-1}\left(\frac{77}{36}\right)=0sin−1(17α​)+cos−1(54​)−tan−1(3677​)=0

with 0<α<130<\alpha<130<α<13.

So,

sin⁡−1(α17)=tan⁡−1(7736)−cos⁡−1(45)\sin^{-1}\left(\frac{\alpha}{17}\right)=\tan^{-1}\left(\frac{77}{36}\right)-\cos^{-1}\left(\frac45\right)sin−1(17α​)=tan−1(3677​)−cos−1(54​)
  1. Evaluate the known inverse trigonometric values

Let

θ=cos⁡−1(45)\theta=\cos^{-1}\left(\frac45\right)θ=cos−1(54​)

Then

cos⁡θ=45,sin⁡θ=35,tan⁡θ=34\cos\theta=\frac45, \quad \sin\theta=\frac35, \quad \tan\theta=\frac34cosθ=54​,sinθ=53​,tanθ=43​

So,

θ=tan⁡−1(34)\theta=\tan^{-1}\left(\frac34\right)θ=tan−1(43​)

Hence,

sin⁡−1(α17)=tan⁡−1(7736)−tan⁡−1(34)\sin^{-1}\left(\frac{\alpha}{17}\right)=\tan^{-1}\left(\frac{77}{36}\right)-\tan^{-1}\left(\frac34\right)sin−1(17α​)=tan−1(3677​)−tan−1(43​)

Now use

tan⁡−1x−tan⁡−1y=tan⁡−1(x−y1+xy)\tan^{-1}x-\tan^{-1}y=\tan^{-1}\left(\frac{x-y}{1+xy}\right)tan−1x−tan−1y=tan−1(1+xyx−y​)

provided the angle lies in the principal range.

Thus,

tan⁡−1(7736)−tan⁡−1(34)=tan⁡−1(7736−341+7736⋅34)\tan^{-1}\left(\frac{77}{36}\right)-\tan^{-1}\left(\frac34\right) =\tan^{-1}\left(\frac{\frac{77}{36}-\frac34}{1+\frac{77}{36}\cdot\frac34}\right)tan−1(3677​)−tan−1(43​)=tan−1(1+3677​⋅43​3677​−43​​)

Compute numerator:

7736−34=7736−2736=5036=2518\frac{77}{36}-\frac34=\frac{77}{36}-\frac{27}{36}=\frac{50}{36}=\frac{25}{18}3677​−43​=3677​−3627​=3650​=1825​

Compute denominator:

1+7736⋅34=1+231144=375144=125481+\frac{77}{36}\cdot\frac34=1+\frac{231}{144}=\frac{375}{144}=\frac{125}{48}1+3677​⋅43​=1+144231​=144375​=48125​

Therefore,

251812548=2518⋅48125=815\frac{\frac{25}{18}}{\frac{125}{48}}=\frac{25}{18}\cdot\frac{48}{125}=\frac{8}{15}48125​1825​​=1825​⋅12548​=158​

So,

sin⁡−1(α17)=tan⁡−1(815)\sin^{-1}\left(\frac{\alpha}{17}\right)=\tan^{-1}\left(\frac{8}{15}\right)sin−1(17α​)=tan−1(158​)
  1. Convert to sine form

Let

ϕ=tan⁡−1(815)\phi=\tan^{-1}\left(\frac{8}{15}\right)ϕ=tan−1(158​)

Then in the right triangle,

  • opposite =8=8=8
  • adjacent =15=15=15
  • hypotenuse =17=17=17

Hence,

sin⁡ϕ=817\sin\phi=\frac{8}{17}sinϕ=178​

Thus,

ϕ=sin⁡−1(817)\phi=\sin^{-1}\left(\frac{8}{17}\right)ϕ=sin−1(178​)

So,

sin⁡−1(α17)=sin⁡−1(817)\sin^{-1}\left(\frac{\alpha}{17}\right)=\sin^{-1}\left(\frac{8}{17}\right)sin−1(17α​)=sin−1(178​)

Since both sides lie in the principal range [−π2,π2]\left[-\frac\pi2,\frac\pi2\right][−2π​,2π​], we get

α17=817\frac{\alpha}{17}=\frac{8}{17}17α​=178​

Therefore,

α=8\alpha=8α=8
  1. Now evaluate sin⁡−1(sin⁡α)+cos⁡−1(cos⁡α)\sin^{-1}(\sin \alpha)+\cos^{-1}(\cos \alpha)sin−1(sinα)+cos−1(cosα)

We need

sin⁡−1(sin⁡8)+cos⁡−1(cos⁡8)\sin^{-1}(\sin 8)+\cos^{-1}(\cos 8)sin−1(sin8)+cos−1(cos8)

where 888 is in radians.

Since

2π<8<3π2\pi<8<3\pi2π<8<3π

because

2π≈6.283,3π≈9.4252\pi\approx 6.283, \qquad 3\pi\approx 9.4252π≈6.283,3π≈9.425

write

8=2π+β,where β=8−2π8=2\pi+\beta, \quad \text{where } \beta=8-2\pi8=2π+β,where β=8−2π

Now

0<β<π0<\beta<\pi0<β<π

and in fact

π2<β<π\frac\pi2<\beta<\pi2π​<β<π

(because 8−2π≈1.717>π28-2\pi\approx1.717>\frac\pi28−2π≈1.717>2π​).

(i) Evaluate sin⁡−1(sin⁡8)\sin^{-1}(\sin 8)sin−1(sin8)

Using periodicity,

sin⁡8=sin⁡(8−2π)=sin⁡β\sin 8=\sin(8-2\pi)=\sin\betasin8=sin(8−2π)=sinβ

Since β∈[π2,π]\beta\in\left[\frac\pi2,\pi\right]β∈[2π​,π],

sin⁡−1(sin⁡β)=π−β\sin^{-1}(\sin\beta)=\pi-\betasin−1(sinβ)=π−β

Hence,

sin⁡−1(sin⁡8)=π−(8−2π)=3π−8\sin^{-1}(\sin 8)=\pi-(8-2\pi)=3\pi-8sin−1(sin8)=π−(8−2π)=3π−8

(ii) Evaluate cos⁡−1(cos⁡8)\cos^{-1}(\cos 8)cos−1(cos8)

Using periodicity,

cos⁡8=cos⁡(8−2π)=cos⁡β\cos 8=\cos(8-2\pi)=\cos\betacos8=cos(8−2π)=cosβ

Since β∈[0,π]\beta\in[0,\pi]β∈[0,π], principal value gives

cos⁡−1(cos⁡β)=β=8−2π\cos^{-1}(\cos\beta)=\beta=8-2\picos−1(cosβ)=β=8−2π

So,

cos⁡−1(cos⁡8)=8−2π\cos^{-1}(\cos 8)=8-2\picos−1(cos8)=8−2π
  1. Add them
sin⁡−1(sin⁡8)+cos⁡−1(cos⁡8)=(3π−8)+(8−2π)=π\sin^{-1}(\sin 8)+\cos^{-1}(\cos 8) =(3\pi-8)+(8-2\pi)=\pisin−1(sin8)+cos−1(cos8)=(3π−8)+(8−2π)=π
  1. Check options

The value is

π\piπ

So the correct option is B.

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