AtomicReaper
  • AtomicReaper
Help please will medal
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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AtomicReaper
  • AtomicReaper
AtomicReaper
  • AtomicReaper
@iTired, Can you help with this?
AtomicReaper
  • AtomicReaper
@MyNameIsNemo

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MyNameIsNemo
  • MyNameIsNemo
Idk that Class XD
Michele_Laino
  • Michele_Laino
try to use this identity: \[\huge \cos \left( {\frac{\pi }{2} - x} \right) = \sin x\]
AtomicReaper
  • AtomicReaper
See I have a problem, I don't know how to work this
Michele_Laino
  • Michele_Laino
If I apply my identity above, I get: \[\Large \frac{{\cos {{\left( {\frac{\pi }{2} - x} \right)}^2}}}{{\sqrt {1 - {{\left( {\sin x} \right)}^2}} }} = \frac{{{{\left( {\sin x} \right)}^2}}}{{\cos x}} = ...?\] since: \[\Large \sqrt {1 - {{\left( {\sin x} \right)}^2}} = \cos x\]
AtomicReaper
  • AtomicReaper
So this probably would have been smart to give in the first place but how would I get it to this?
Michele_Laino
  • Michele_Laino
you have to apply this identity: \[\Large \frac{{\sin x}}{{\cos x}} = \tan x\] please keep in mind that we can write this: \[\Large \frac{{\cos {{\left( {\frac{\pi }{2} - x} \right)}^2}}}{{\sqrt {1 - {{\left( {\sin x} \right)}^2}} }} = \frac{{{{\left( {\sin x} \right)}^2}}}{{\cos x}} = \frac{{\sin x}}{{\cos x}} \cdot \sin x = ...?\]

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