eroshea
x=(tanA+cotA)^2 sinAtan^2A



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henpen
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\[x=(tanA+cotA)^2 *(sinA\tan^2A)\]This?

uzumakhi
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is @henpen right?

eroshea
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yes

henpen
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\[(tanA+cotA)=(\frac{sinA}{cosA}+\frac{cosA}{sinA})=(\frac{sinAsinA}{cosAsinA}+\frac{cosAcosA}{sinAcosA})=\frac{1}{cosAsinA}\]

henpen
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\[\because \cos^2(x)+\sin^2(x)=1\]

henpen
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\[sinA\tan^2A=\frac{sinAcos^2A}{\cos^2A}\frac{\sin^2A}{\cos^2A}=\frac{sinAcos^2A\sin^2A}{\cos^2A}\]

eroshea
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give me a minute to analyze what you had illustrated

henpen
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Multiplying,\[(\frac{1}{cosAsinA})^2*(\frac{sinAcos^2A\sin^2A}{\cos^2A})=\frac{sinAcos^2A\sin^2A}{\cos^4Asin^2A}\]

henpen
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Which you can simplify if you want.

Zekarias
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you must get this csc(x)sec^2(x)

eroshea
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owh.. i'm sorry .. you should find the value of x?
the possible answers are:
a. 4 b.3 c.2 d.1

henpen
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\[cscAsec^2A\sec^4A\]I must be mistaken.

eroshea
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how will i arrive at the possible answers?