Question

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Answer 1

$\frac{tanA}{1 - cotA} + \frac{cotA}{1 -tanA}$

$\frac{ \frac{sinA}{cosA} }{1 - \frac{cosA}{sinA} } + \frac{ \frac{cosA}{sinA} }{1 - \frac{sinA}{cosA} }$

$\frac{ \frac{sinA}{cosA} }{ \frac{sinA - cosA}{sinA} } + \frac{ \frac{cosA}{sinA} }{ \frac{cosA - sinA}{cosA} }$

$\frac{sin^2A}{cosA(sinA - cosA) } + \frac{cos^2A}{sinA(cosA - sinA)}$

$[ \frac{sin^2A}{cosA(sinA - cosA)}] - [\frac{cos^2A}{sinA(sinA - cosA)}]$

$\frac{1}{sinA - cosA} [ { \frac{sin^2A}{cosA} - \frac{cos^2A}{sinA}]$

$\frac{1}{sinA - cosA}$×$\frac{sin^3A - cos^3A}{cosA sinA}$

$\frac{(sinA - cosA)(sin^2A + sinAcosA + cos^2A)}{(sinA - cosA)(sinAcosA)}$

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Then,

For, 1 + tanФ + cotФ 

⇒ $\frac{sin^2A + sinAcosA + cos^2A}{sinAcosA}$

⇒$\frac{sin^2A}{sinA cosA} + \frac{sinAcosA}{sinAcosA} + \frac{cos^2A}{sinAcosA}$

⇒$\frac{sinA}{cosA} + 1 + \frac{cosA}{sinA}$

⇒ tanA + 1  + cotA

              Hence, Proved                              


For, 1 + secФ cosecФ

⇒ $\frac{sin^2A + cos^2A + sinAcosA}{sinAcosA}$

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We know, sin²A + cos²A = 1 |
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⇒$\frac{1}{sinAcosA} + \frac{sinA cosA}{sinAcosA}$

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1/sin = cosec
1/cos = sec
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⇒ cosecA secA + 1 

                               Hence, Proved                     





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