SOLUTION: How do I find {{{ cot( sin^-1(-1/2) - sec^-1(2) ) }}}? I have {{{ x = sin^-1(-1/2)}}} implies sin x= -1/2 and {{{ y = sec^-1(2)}}} implies sec y= 2. But I'm not sure if co

Algebra ->  Trigonometry-basics -> SOLUTION: How do I find {{{ cot( sin^-1(-1/2) - sec^-1(2) ) }}}? I have {{{ x = sin^-1(-1/2)}}} implies sin x= -1/2 and {{{ y = sec^-1(2)}}} implies sec y= 2. But I'm not sure if co      Log On


   

Question 888015: How do I find +cot%28+sin%5E-1%28-1%2F2%29+-+sec%5E-1%282%29+%29+?
I have +x+=+sin%5E-1%28-1%2F2%29 implies sin x= -1/2
and +y+=+sec%5E-1%282%29 implies sec y= 2.
But I'm not sure if cot( x - y) is the right way to go.
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
cot( sin^-1(-1/2) - sec^-1(2) )
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If sin(x) = -1/2, x = 210 degrees
If sec(y) = 2, y = 60 degrees
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cot(210-60) = cot(50) = 0.8391
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Cheers,
Stan H.
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