prove that :
[tex]cot \alpha + tan \alpha = cosec \alpha \times sec \alpha [/tex]
​

1 thought on “prove that :<br />[tex]cot \alpha + tan \alpha = cosec \alpha \times sec \alpha [/tex]<br />​”

1. [tex]\huge\mathfrak\red{Given:-}[/tex]

   [tex]cot \alpha + tan \alpha = cosec \alpha \times sec \alpha [/tex]

   [tex]\huge\mathcal\blue{Solution:-}[/tex]

   [tex]\huge\mathcal\red{Iñ \: LHS:-}[/tex]

   ➡[tex]cot \alpha + tan \alpha[/tex]

   ➡[tex]cot \alpha = \text{$\frac{cos \alpha}{sin \alpha}$}[/tex]

   ➡[tex]tan \alpha = \text{$\frac{sin \alpha}{cos \alpha}$}[/tex]

   ➡[tex]\text{$\frac{cos^2 \alpha + sin^2 \alpha}{sin \alpha cos \alpha}$}[/tex]

   ➡[tex]cos ^2\alpha + sin^2\alpha = 1[/tex]

   ➡[tex]\text{$\frac{1}{cos \alpha ×sin \alpha}$}[/tex]

   ➡[tex]\text{$\frac{1}{cos \alpha}$}×\text{$\frac{1}{sin \alpha}$}[/tex]

   ➡[tex]\text{$\frac{1}{sin \alpha}$} = cosec \alpha[/tex]

   and ;

   ➡[tex]\text{$\frac{1}{cos \alpha}$} = sec \alpha[/tex]

   So,

   ➡[tex]cosec \alpha × sec \alpha[/tex]

   [tex]\mathcal\red{LHS \: is \: equal \: to \: RHS}[/tex]

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