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Inverse Trigonometric Functions question

2022 · 30 Jun · Shift 1 · Q33
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  5. /2022 · 30 Jun · Shift 1 · Q33

Inverse Trigonometric Functions question

2022 · 30 Jun · Shift 1 · Q33

JEE MainMathematicsInverse Trigonometric FunctionsMCQ+4 / −1
Let α=tan⁡(5π16sin⁡(2cos⁡−1(15)))\alpha = \tan \left( {{{5\pi } \over {16}}\sin \left( {2{{\cos }^{ - 1}}\left( {{1 \over {\sqrt 5 }}} \right)} \right)} \right)α=tan(165π​sin(2cos−1(5​1​))) and β=cos⁡(sin⁡−1(45)+sec⁡−1(53))\beta = \cos \left( {{{\sin }^{ - 1}}\left( {{4 \over 5}} \right) + {{\sec }^{ - 1}}\left( {{5 \over 3}} \right)} \right)β=cos(sin−1(54​)+sec−1(35​)) where the inverse trigonometric functions take principal values. Then, the equation whose roots are α\alphaα and β\betaβ is :
  1. A
    15x2−8x−7=015{x^2} - 8x - 7 = 015x2−8x−7=0
  2. B
    5x2−12x+7=05{x^2} - 12x + 7 = 05x2−12x+7=0
  3. C
    25x2−18x−7=025{x^2} - 18x - 7 = 025x2−18x−7=0
  4. D
    25x2−32x+7=025{x^2} - 32x + 7 = 025x2−32x+7=0
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