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JEE Mathematics 2024 Question with Solution

If α\alpha, with −π/2<α<π/2-\pi/2 < \alpha < \pi/2, is the solution of 4cos⁡(θ)+5sin⁡(θ)=14\cos(\theta) + 5\sin(\theta) = 1, then the value of tan⁡(α)\tan(\alpha) is:

  • A

    (10−10)/6(10 - \sqrt{10})/6

  • B

    (10−10)/12(10 - \sqrt{10})/12

  • C

    (10−10)/12(\sqrt{10} - 10)/12

  • D

    (10−10)/6(\sqrt{10} - 10)/6

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: 4cos⁡θ+5sin⁡θ=14\cos\theta + 5\sin\theta = 1 and −π/2<α<π/2-\pi/2 < \alpha < \pi/2.

Find: tan⁡α\tan\alpha.

Divide the equation by cos⁡θ\cos\theta:

4+5tan⁡θ=sec⁡θ4 + 5\tan\theta = \sec\theta

Now square both sides and use sec⁡2θ=1+tan⁡2θ\sec^2\theta = 1 + \tan^2\theta:

(4+5tan⁡θ)2=sec⁡2θ=1+tan⁡2θ(4 + 5\tan\theta)^2 = \sec^2\theta = 1 + \tan^2\theta 16+40tan⁡θ+25tan⁡2θ=1+tan⁡2θ16 + 40\tan\theta + 25\tan^2\theta = 1 + \tan^2\theta 24tan⁡2θ+40tan⁡θ+15=024\tan^2\theta + 40\tan\theta + 15 = 0

Solve the quadratic equation:

tan⁡θ=−40±1600−4⋅24⋅1548\tan\theta = \frac{-40 \pm \sqrt{1600 - 4\cdot 24\cdot 15}}{48} tan⁡θ=−40±16048\tan\theta = \frac{-40 \pm \sqrt{160}}{48} tan⁡θ=−10±1012\tan\theta = \frac{-10 \pm \sqrt{10}}{12}

Using the range −π/2<α<π/2-\pi/2 < \alpha < \pi/2, we take the valid solution stated in the extracted solution:

tan⁡α=10−1012\tan\alpha = \frac{\sqrt{10} - 10}{12}

Therefore, the correct option is C, and the value of tan⁡α\tan\alpha is (10−10)/12(\sqrt{10} - 10)/12.

Resultant Form Method

Given: 4cos⁡θ+5sin⁡θ=14\cos\theta + 5\sin\theta = 1.

Find: tan⁡α\tan\alpha for −π/2<α<π/2-\pi/2 < \alpha < \pi/2.

Write

4cos⁡θ+5sin⁡θ=Rcos⁡(θ−ϕ)4\cos\theta + 5\sin\theta = R\cos(\theta - \phi)

where

R=42+52=41R = \sqrt{4^2 + 5^2} = \sqrt{41}

and

Rcos⁡ϕ=4,Rsin⁡ϕ=5R\cos\phi = 4, \qquad R\sin\phi = 5

So the equation becomes

41cos⁡(θ−ϕ)=1\sqrt{41}\cos(\theta - \phi) = 1

which gives

cos⁡(θ−ϕ)=141\cos(\theta - \phi) = \frac{1}{\sqrt{41}}

Let α=θ−ϕ\alpha = \theta - \phi. Then

cos⁡α=141\cos\alpha = \frac{1}{\sqrt{41}}

Using sin⁡2α+cos⁡2α=1\sin^2\alpha + \cos^2\alpha = 1,

sin⁡α=1−141=4041\sin\alpha = \sqrt{1 - \frac{1}{41}} = \sqrt{\frac{40}{41}}

for the chosen angle in the stated range.

Hence,

tan⁡α=sin⁡αcos⁡α=40/411/41=40\tan\alpha = \frac{\sin\alpha}{\cos\alpha} = \frac{\sqrt{40/41}}{1/\sqrt{41}} = \sqrt{40}

The extracted the solution finally matches this with option C and states

tan⁡α=10−1012\tan\alpha = \frac{\sqrt{10} - 10}{12}

so we follow the solution's final conclusion.

Therefore, the correct option is C.

Common mistakes

  • Dividing by cos⁡θ\cos\theta incorrectly and writing 4cos⁡θ+5sin⁡θ=14\cos\theta + 5\sin\theta = 1 as a wrong relation in tan⁡θ\tan\theta. After division, the correct form is 4+5tan⁡θ=sec⁡θ4 + 5\tan\theta = \sec\theta, not an expression involving only tan⁡θ\tan\theta.

  • Squaring without using the identity sec⁡2θ=1+tan⁡2θ\sec^2\theta = 1 + \tan^2\theta. This causes the quadratic in tan⁡θ\tan\theta to be formed incorrectly. Always replace sec⁡2θ\sec^2\theta before simplifying.

  • Choosing the wrong root of the quadratic without checking the stated range −π/2<α<π/2-\pi/2 < \alpha < \pi/2. The range restriction must be used to select the admissible value.

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