Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Trigonometric Functions and Equations question

2024 · 29 Jan · Shift 1 · Q34
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Mathematics
  4. /Trigonometric Functions and Equations
  5. /2024 · 29 Jan · Shift 1 · Q34

Trigonometric Functions and Equations question

2024 · 29 Jan · Shift 1 · Q34

JEE MainMathematicsTrigonometric Functions and EquationsMCQ+4 / −1
If α,−π2<α<π2\alpha,-\frac{\pi}{2}\lt \alpha\lt \frac{\pi}{2}α,−2π​<α<2π​ is the solution of 4cos⁡θ+5sin⁡θ=14 \cos \theta+5 \sin \theta=14cosθ+5sinθ=1, then the value of tan⁡α\tan \alphatanα is
  1. A
    10−1012\frac{10-\sqrt{10}}{12}1210−10​​
  2. B
    10−106\frac{\sqrt{10}-10}{6}610​−10​
  3. C
    10−1012\frac{\sqrt{10}-10}{12}1210​−10​
  4. D
    10−106\frac{10-\sqrt{10}}{6}610−10​​
View written solutionFree

Correct answer: C

  1. Interpret the equation

We are to solve 4cos⁡α+5sin⁡α=1,4\cos\alpha + 5\sin\alpha = 1,4cosα+5sinα=1, with −π2<α<π2.-\frac{\pi}{2} < \alpha < \frac{\pi}{2}.−2π​<α<2π​.

We need to find tan⁡α\tan\alphatanα.

  1. Use the identity sin⁡2α+cos⁡2α=1\sin^2\alpha + \cos^2\alpha = 1sin2α+cos2α=1

Let sin⁡α=s,cos⁡α=c.\sin\alpha = s, \qquad \cos\alpha = c.sinα=s,cosα=c. Then 4c+5s=1.4c + 5s = 1.4c+5s=1. So, c=1−5s4.c = \frac{1-5s}{4}.c=41−5s​.

Now use s2+c2=1.s^2 + c^2 = 1.s2+c2=1. Substituting ccc: s2+(1−5s4)2=1.s^2 + \left(\frac{1-5s}{4}\right)^2 = 1.s2+(41−5s​)2=1.

Multiply by 161616: 16s2+(1−5s)2=16.16s^2 + (1-5s)^2 = 16.16s2+(1−5s)2=16.

Expand: 16s2+1−10s+25s2=16.16s^2 + 1 - 10s + 25s^2 = 16.16s2+1−10s+25s2=16. 41s2−10s−15=0.41s^2 - 10s - 15 = 0.41s2−10s−15=0.

Solve this quadratic: s=10±100+4⋅41⋅1582s = \frac{10 \pm \sqrt{100 + 4\cdot 41\cdot 15}}{82}s=8210±100+4⋅41⋅15​​ =10±256082= \frac{10 \pm \sqrt{2560}}{82}=8210±2560​​ =10±161082= \frac{10 \pm 16\sqrt{10}}{82}=8210±1610​​ =5±81041.= \frac{5 \pm 8\sqrt{10}}{41}.=415±810​​.

So, sin⁡α=5±81041.\sin\alpha = \frac{5 \pm 8\sqrt{10}}{41}.sinα=415±810​​.

  1. Find the corresponding values of cos⁡α\cos\alphacosα

Using c=1−5s4,c = \frac{1-5s}{4},c=41−5s​, for s=5+81041,s = \frac{5 + 8\sqrt{10}}{41},s=415+810​​, we get

= \frac{\frac{41-25-40\sqrt{10}}{41}}{4} = \frac{16-40\sqrt{10}}{164} = \frac{4-10\sqrt{10}}{41},$$ which is negative. For $$s = \frac{5 - 8\sqrt{10}}{41},$$ we get $$c = \frac{1 - 5\cdot \frac{5-8\sqrt{10}}{41}}{4} = \frac{\frac{41-25+40\sqrt{10}}{41}}{4} = \frac{16+40\sqrt{10}}{164} = \frac{4+10\sqrt{10}}{41},$$ which is positive. Since $$-\frac{\pi}{2} < \alpha < \frac{\pi}{2},$$ we must have $$\cos\alpha > 0.$$ Therefore, $$\sin\alpha = \frac{5 - 8\sqrt{10}}{41}, \qquad \cos\alpha = \frac{4 + 10\sqrt{10}}{41}.$$ --- 4. **Compute $\tan\alpha$** $$\tan\alpha = \frac{\sin\alpha}{\cos\alpha} = \frac{\frac{5 - 8\sqrt{10}}{41}}{\frac{4 + 10\sqrt{10}}{41}} = \frac{5 - 8\sqrt{10}}{4 + 10\sqrt{10}}.$$ Rationalize: $$\tan\alpha = \frac{(5 - 8\sqrt{10})(4 - 10\sqrt{10})}{(4 + 10\sqrt{10})(4 - 10\sqrt{10})}.$$ Denominator: $$16 - 1000 = -984.$$ Numerator: $$20 - 50\sqrt{10} - 32\sqrt{10} + 800 = 820 - 82\sqrt{10}.

Thus,

= -\frac{10 - \sqrt{10}}{12} = \frac{\sqrt{10} - 10}{12}.$$ --- 5. **Match with the options** $$\tan\alpha = \frac{\sqrt{10}-10}{12}.$$ This is **Option C**. --- 6. **Comparison with stored answer** Stored correct answer: **C** Our derived answer: **C** So they agree.
PreviousNext

More from Trigonometric Functions and Equations

  • The sum of the solutions x∈R of the equation cos6x−sin6x3cos2x+cos32x​=x3−x2+6 is2024 · MCQ
  • If 2sin3x+sin2xcosx+4sinx−4=0 has exactly 3 solutions in the interval [0,2nπ​],n∈N, then the roots of the equation x2+nx+(n−3)=0 belong to :2024 · MCQ
  • If θ∈[−2π,2π], then the number of solutions of 22​cos2θ+(2−6​)cosθ−3​=0, is equal to:2025 · MCQ
  • If θϵ[−67π​,34π​], then the number of solutions of 3​cosec2θ−2(3​−1)cosecθ−4=0, is equal to :2025 · MCQ
  •  The number of solutions of the equation 2x+3tanx=π,x∈[−2π,2π]−{±2π​,±23π​} is: 2025 · MCQ
  • The number of solutions of the equation (4−3​)sinx−23​cos2x=−1+3​4​,x∈[−2π,25π​] is2025 · MCQ
  • The number of solutions of the equation cos2θcos2θ​+cos25θ​=2cos325θ​ in [−2π​,2π​] is :2025 · MCQ
  • The sum of all values of θ∈[0,2π] satisfying 2sin2θ=cos2θ and 2cos2θ=3sinθ is2025 · MCQ