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JEE Main 201815 AprilMorning ShiftOnline
Exam
JEE Main
Year
2018
Date
15 April
Shift
Morning Shift
Mode
Online
Subject
Mathematics
Chapter
Trigonometric Equations

JEE Main 2018 · 15 April · Morning Shift · Online · Mathematics · Trigonometric Equations

If A and B are the roots of the quadratic equation, 3 x^2-10 x-25=0...

If tan⁡A\tan A and tan⁡B\tan B are the roots of the quadratic equation, 3x2−10x−25=03 x^2-10 x-25=0 then the value of 3sin⁡2(A+B)−10sin⁡(A+B)⋅cos⁡(A+B)−25cos⁡23 \sin ^2(A+B)-10 \sin (A+B) \cdot \cos (A+B)-25 \cos ^2 (A+B)(A+B) is

Options

  1. A25
  2. B−25-25
  3. C−10-10
  4. D10

Correct answer

B. −25-25

Step-by-step solution

As tan⁡A\tan A and tan⁡B\tan B are the roots of 3x2−10x−25=03 x^2-10 x-25=0 So, tan⁡(A+B)=tan⁡A+tan⁡B1−tan⁡Atan⁡B=1031+253\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A \tan B}=\frac{\frac{10}{3}}{1+\frac{25}{3}} =10/328/3=514=\frac{10 / 3}{28 / 3}=\frac{5}{14} Now, cos⁡2(A+B)=−1+2cos⁡2(A+B)\cos ^2(A+B)=-1+2 \cos ^2(A+B) =1−tan⁡2(A+B)1+tan⁡2(A+B)⇒cos⁡2(A+B)=196221=\frac{1-\tan ^2(A+B)}{1+\tan ^2(A+B)} \Rightarrow \cos ^2(A+B)=\frac{196}{221} ∴3sin⁡2(A+B)−10sin⁡(A+B)\therefore 3 \sin ^2(A+B)-10 \sin (A+B) =cos⁡2(A+B)=\cos ^2(A+B)\left[3 \tan ^2(A+B)-10 \tan (A+B)-25\right] =\frac{75-700-4900}{196} \times \frac{196}{221}=-\frac{5525}{196} \times \frac{196}{221}=-25
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