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JEE Main 20245 AprilMorning Shift
Exam
JEE Main
Year
2024
Date
5 April
Shift
Morning Shift
Subject
Mathematics
Chapter
Trigonometric Equations

JEE Main 2024 · 5 April · Morning Shift · Mathematics · Trigonometric Equations

Suppose θ [0, (π)/(4)] is a solution of 4 θ-3 θ=1 . Then θ is equal...

Suppose θϵ[0,π4]\theta \epsilon\left[0, \frac{\pi}{4}\right] is a solution of 4cos⁡θ−3sin⁡θ=14 \cos \theta-3 \sin \theta=1. Then cos⁡θ\cos \theta is equal to :

Options

  1. A4(36+2)\frac{4}{(3 \sqrt{6}+2)}
  2. B6+6(36+2)\frac{6+\sqrt{6}}{(3 \sqrt{6}+2)}
  3. C4(36−2)\frac{4}{(3 \sqrt{6}-2)}
  4. D6−6(36−2)\frac{6-\sqrt{6}}{(3 \sqrt{6}-2)}

Correct answer

C. 4(36−2)\frac{4}{(3 \sqrt{6}-2)}

Step-by-step solution

44\left(\frac{1-\tan ^2 \theta / 2}{1+\tan ^2 \theta / 2}\right)-3(2tan⁡θ21+tan⁡2θ2)=1\left(\frac{2 \tan \frac{\theta}{2}}{1+\tan ^2 \frac{\theta}{2}}\right)=1 let tan⁡θ2=t\tan \frac{\theta}{2}=\mathrm{t}4−4t2−6t1+t2=14−4t2−6t=1+t2\begin{aligned} & \frac{4-4 t^2-6 t}{1+t^2}=1 \\ & 4-4 t^2-6 t=1+t^2 \end{aligned}⇒5t2+6t−3=0⇒t=−6±36−4(5)(−3)2(5)=−6±9610=−6±4610t=−3+265cos⁡θ=1−t21+t2=1−(26−35)21+(26−35)2=1−(24+9−12625)1+(24+9−12625)=25−33+12625+33−126=126−858−126=66−429−66×29+6629+66=100+1506625=4+6625×4−664−66=−20025(4−66)=−84−66=436−2\begin{aligned} & \Rightarrow 5 t^2+6 t-3=0 \\ & \Rightarrow t=\frac{-6 \pm \sqrt{36-4(5)(-3)}}{2(5)} \\ & =\frac{-6 \pm \sqrt{96}}{10} \\ & =\frac{-6 \pm 4 \sqrt{6}}{10} \\ & t=\frac{-3+2 \sqrt{6}}{5} \\ & \cos \theta=\frac{1-t^2}{1+t^2}=\frac{1-\left(\frac{2 \sqrt{6}-3}{5}\right)^2}{1+\left(\frac{2 \sqrt{6}-3}{5}\right)^2}=\frac{1-\left(\frac{24+9-12 \sqrt{6}}{25}\right)}{1+\left(\frac{24+9-12 \sqrt{6}}{25}\right)} \\ & =\frac{25-33+12 \sqrt{6}}{25+33-12 \sqrt{6}}=\frac{12 \sqrt{6}-8}{58-12 \sqrt{6}}=\frac{6 \sqrt{6}-4}{29-6 \sqrt{6}} \times \frac{29+6 \sqrt{6}}{29+6 \sqrt{6}} \\ & =\frac{100+150 \sqrt{6}}{625}=\frac{4+6 \sqrt{6}}{25} \times \frac{4-6 \sqrt{6}}{4-6 \sqrt{6}} \\ & =\frac{-200}{25(4-6 \sqrt{6})}=\frac{-8}{4-6 \sqrt{6}}=\frac{4}{3 \sqrt{6}-2}\end{aligned}$$
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