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Inverse Trigonometric Functions question

2022 · 30 Jun · Shift 1 · Q33
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  5. /2022 · 30 Jun · Shift 1 · Q33

Inverse Trigonometric Functions question

2022 · 30 Jun · Shift 1 · Q33

JEE MainMathematicsInverse Trigonometric FunctionsMCQ+4 / −1
Let α=tan⁡(5π16sin⁡(2cos⁡−1(15)))\alpha = \tan \left( {{{5\pi } \over {16}}\sin \left( {2{{\cos }^{ - 1}}\left( {{1 \over {\sqrt 5 }}} \right)} \right)} \right)α=tan(165π​sin(2cos−1(5​1​))) and β=cos⁡(sin⁡−1(45)+sec⁡−1(53))\beta = \cos \left( {{{\sin }^{ - 1}}\left( {{4 \over 5}} \right) + {{\sec }^{ - 1}}\left( {{5 \over 3}} \right)} \right)β=cos(sin−1(54​)+sec−1(35​)) where the inverse trigonometric functions take principal values. Then, the equation whose roots are α\alphaα and β\betaβ is :
  1. A
    15x2−8x−7=015{x^2} - 8x - 7 = 015x2−8x−7=0
  2. B
    5x2−12x+7=05{x^2} - 12x + 7 = 05x2−12x+7=0
  3. C
    25x2−18x−7=025{x^2} - 18x - 7 = 025x2−18x−7=0
  4. D
    25x2−32x+7=025{x^2} - 32x + 7 = 025x2−32x+7=0
View written solutionFree

Correct answer: C

  1. Find α\alphaα

Given

α=tan⁡(5π16sin⁡(2cos⁡−1(15)))\alpha = \tan\left(\frac{5\pi}{16}\sin\left(2\cos^{-1}\left(\frac1{\sqrt5}\right)\right)\right)α=tan(165π​sin(2cos−1(5​1​)))

Let

θ=cos⁡−1(15)\theta = \cos^{-1}\left(\frac1{\sqrt5}\right)θ=cos−1(5​1​)

Then

cos⁡θ=15\cos\theta = \frac1{\sqrt5}cosθ=5​1​

Since θ\thetaθ is a principal value of cos⁡−1\cos^{-1}cos−1, θ∈[0,π]\theta\in[0,\pi]θ∈[0,π], and here θ\thetaθ is acute. So

sin⁡θ=1−15=25\sin\theta = \sqrt{1-\frac15} = \frac{2}{\sqrt5}sinθ=1−51​​=5​2​

Now,

sin⁡2θ=2sin⁡θcos⁡θ=2⋅25⋅15=45\sin 2\theta = 2\sin\theta\cos\theta = 2\cdot \frac{2}{\sqrt5}\cdot \frac{1}{\sqrt5} = \frac{4}{5}sin2θ=2sinθcosθ=2⋅5​2​⋅5​1​=54​

Hence

α=tan⁡(5π16⋅45)=tan⁡(π4)=1\alpha = \tan\left(\frac{5\pi}{16}\cdot \frac45\right)=\tan\left(\frac{\pi}{4}\right)=1α=tan(165π​⋅54​)=tan(4π​)=1
  1. Find β\betaβ

Given

β=cos⁡(sin⁡−1(45)+sec⁡−1(53))\beta = \cos\left(\sin^{-1}\left(\frac45\right)+\sec^{-1}\left(\frac53\right)\right)β=cos(sin−1(54​)+sec−1(35​))

Let

A=sin⁡−1(45),B=sec⁡−1(53)A=\sin^{-1}\left(\frac45\right), \qquad B=\sec^{-1}\left(\frac53\right)A=sin−1(54​),B=sec−1(35​)

Then

sin⁡A=45\sin A = \frac45sinA=54​

Since A∈[−π2,π2]A\in\left[-\frac\pi2,\frac\pi2\right]A∈[−2π​,2π​], we get AAA acute, so

cos⁡A=35\cos A = \frac35cosA=53​

Also,

sec⁡B=53  ⟹  cos⁡B=35\sec B = \frac53 \implies \cos B = \frac35secB=35​⟹cosB=53​

For principal value of sec⁡−1\sec^{-1}sec−1, BBB is acute here, so

sin⁡B=45\sin B = \frac45sinB=54​

Now use

cos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡B\cos(A+B)=\cos A\cos B - \sin A\sin Bcos(A+B)=cosAcosB−sinAsinB

Thus

β=35⋅35−45⋅45\beta = \frac35\cdot\frac35 - \frac45\cdot\frac45β=53​⋅53​−54​⋅54​ β=925−1625=−725\beta = \frac{9}{25}-\frac{16}{25}=-\frac{7}{25}β=259​−2516​=−257​
  1. Form the quadratic equation with roots α\alphaα and β\betaβ

Roots are

α=1,β=−725\alpha=1, \qquad \beta=-\frac{7}{25}α=1,β=−257​

So the equation is

x2−(α+β)x+αβ=0x^2-(\alpha+\beta)x+\alpha\beta=0x2−(α+β)x+αβ=0

Compute sum:

α+β=1−725=1825\alpha+\beta = 1-\frac{7}{25}=\frac{18}{25}α+β=1−257​=2518​

Compute product:

αβ=1⋅(−725)=−725\alpha\beta = 1\cdot\left(-\frac{7}{25}\right)=-\frac{7}{25}αβ=1⋅(−257​)=−257​

Hence

x2−1825x−725=0x^2-\frac{18}{25}x-\frac{7}{25}=0x2−2518​x−257​=0

Multiplying by 252525,

25x2−18x−7=025x^2-18x-7=025x2−18x−7=0
  1. Match with options

This is Option C.

  1. Comparison with stored correct answer

Stored correct answer: C

Our derived answer: C

So they agree.

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