sin(sin\(\frac{2π}{3}\)) =
  • \(\frac{2π}{3}\)
  • \(\frac{π}{6}\)
  • \(\frac{4π}{3}\)
  • \(\frac{π}{3}\)
sin(1 – x) – 2 sinx = \(\frac{π}{2}\) then x = ?
  • 0, \(\frac{1}{2}\)
  • 1, \(\frac{1}{2}\)
  • \(\frac{1}{2}\)
  • 0
tan√3 – sec(-2)
  • π
  • –\(\frac{π}{3}\), 0
  • \(\frac{π}{3}\)
  • \(\frac{2π}{3}\)
sin(secx + cosecx) =
  • 1
  • -1
  • \(\frac{π}{2}\)
  • \(\frac{π}{3}\)
The principle value of sin\(\frac{√3}{2}\) is
  • \(\frac{2π}{3}\)
  • \(\frac{π}{6}\)
  • \(\frac{π}{4}\)
  • \(\frac{π}{3}\)
Simplified form of cos(4x– 3x)
  • 3 sinx
  • 3 cosx
  • π – 3 sinx
  • None of these
The value of tan(tan\(\frac{4}{5}\) + tan\(\frac{2}{3}\)) is
  • \(\frac{6}{17}\)
  • \(\frac{7}{16}\)
  • \(\frac{17}{6}\)
  • None of these
The value of x for which sin |cot
  • \(\frac{2}{1}\)
  • 1
  • 0
  • \(\frac{1}{2}\)
Princal value of cos(\(\frac{-1}{√2}\))
  • \(\frac{3π}{4}\)
  • \(\frac{5π}{4}\)
  • –\(\frac{π}{4}\)
  • None of these
tan √3 – sec(-2) is equal to
  • π
  • –\(\frac{π}{3}\)
  • \(\frac{π}{3}\)
  • \(\frac{2π}{3}\)
If y = sec x then
  • 0 ≤ y ≤ π
  • 0 ≤ y ≤ \(\frac{π}{2}\)
  • –\(\frac{π}{2}\) < y < \(\frac{π}{2}\)
  • None of these
If x + \(\frac{1}{x}\) = 2 then the principal value of sin x is x
  • \(\frac{π}{4}\)
  • \(\frac{π}{2}\)
  • π
  • \(\frac{3π}{2}\)
The principle value of sin(sin\(\frac{2π}{3}\)) is
  • \(\frac{2π}{3}\)
  • \(\frac{π}{3}\)
  • \(\frac{-π}{6}\)
  • \(\frac{π}{6}\)
Algebraic expression for sin (cot x) is
  • \(\frac{1}{1+x^2}\)
  • \(\frac{1}{\sqrt{1+x^2}}\)
  • \(\frac{x}{\sqrt{1+x^2}}\)
  • None of these
Princal value of tan (-1) is
  • \(\frac{π}{4}\)
  • \(\frac{-π}{2}\)
  • \(\frac{5π}{4}\)
  • \(\frac{-π}{4}\)
Principal value of sin(\(\frac{1}{√2}\))
  • \(\frac{π}{4}\)
  • \(\frac{3π}{4}\)
  • \(\frac{5π}{4}\)
  • None of these
sin x = y Then
  • 0 ≤ y ≤ π
  • –\(\frac{π}{2}\) ≤ y ≤ \(\frac{π}{2}\)
  • 0 < y < π
  • –\(\frac{π}{2}\) < y < –\(\frac{π}{2}\)
cos(cos\(\frac{7π}{6}\)) is equal to
  • \(\frac{7π}{6}\)
  • \(\frac{5π}{6}\)
  • \(\frac{π}{3}\)
  • \(\frac{π}{6}\)
sin[\(\frac{π}{3}\) – sin(-\(\frac{1}{2}\))] is equal to
  • \(\frac{1}{2}\)
  • \(\frac{1}{3}\)
  • \(\frac{1}{4}\)
  • 1
tan\(\frac{1}{2}\) + tan\(\frac{2}{11}\) = tan a then a = ?
  • \(\frac{1}{4}\)
  • \(\frac{1}{2}\)
  • \(\frac{3}{4}\)
  • 1
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