anonymous
  • anonymous
x=(tanA+cotA)^2 sinA-tan^2A
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
\[x=(tanA+cotA)^2 *(sinA-\tan^2A)\]This?
anonymous
  • anonymous
is @henpen right?
anonymous
  • anonymous
yes

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anonymous
  • anonymous
\[(tanA+cotA)=(\frac{sinA}{cosA}+\frac{cosA}{sinA})=(\frac{sinAsinA}{cosAsinA}+\frac{cosAcosA}{sinAcosA})=\frac{1}{cosAsinA}\]
anonymous
  • anonymous
\[\because \cos^2(x)+\sin^2(x)=1\]
anonymous
  • anonymous
\[sinA-\tan^2A=\frac{sinAcos^2A}{\cos^2A}-\frac{\sin^2A}{\cos^2A}=\frac{sinAcos^2A-\sin^2A}{\cos^2A}\]
anonymous
  • anonymous
give me a minute to analyze what you had illustrated
anonymous
  • anonymous
Multiplying,\[(\frac{1}{cosAsinA})^2*(\frac{sinAcos^2A-\sin^2A}{\cos^2A})=\frac{sinAcos^2A-\sin^2A}{\cos^4Asin^2A}\]
anonymous
  • anonymous
Which you can simplify if you want.
anonymous
  • anonymous
you must get this csc(x)-sec^2(x)
anonymous
  • anonymous
owh.. i'm sorry .. you should find the value of x? the possible answers are: a. 4 b.3 c.2 d.1
anonymous
  • anonymous
\[cscAsec^2A-\sec^4A\]I must be mistaken.
anonymous
  • anonymous
how will i arrive at the possible answers?

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