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JEE Advanced · Mathematics · 8. Trigonometric Equations

For any y, let cot-1y0,π and tan-1y-π2,π2. Then the sum of all the solutions of the equation tan-16y9-y2+cot-19-y26y=2π3 for 0<|y|<3, is equal to

  1. A 23-3
  2. B 3-23
  3. C 43-6
  4. D 6-43
Verified Solution

Answer & Solution

Correct Answer

(C) 43-6

Step-by-step Solution

Detailed explanation

Given,
tan-16y9-y2+cot-19-y26y=2π3
And 0<|y|<3y-3,3-0
Now taking, Case-l:
When 6y9-y2>0y>0
tan16y9y2+tan16y9y2=2π3
2tan-16y9-y2=2π3
tan-16y9-y2=π3
6y9-y2=3
6y=93-3y2
3y2+6y-93=0
3y2+9y-3y-93=0
y+333y-3=0
y-33  y=3 as y(0,3)
Now taking, Case-II: 
When 6y9-y2<0y<0
tan-16y9-y2+π+tan-16y9-y2=2π3
2tan-16y9-y2=-π3
tan-16y9-y2=-π6
6y9-y2=-13
63y=-9+y2
y2-63y-9=0
y=63±108+362=63±122=33±6
 As y(3.0), so y=336
 Sum of solutions =3+(336)=436
From JEE Advanced
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