Question

1 - cos A/ sin A is equal to?

Answer 1

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$= > 1- \frac{ cos \: a}{sin \: }$

$= > 1- \frac{ \frac{b}{h} }{ \frac{p}{h} }$

$= > 1 - \frac{p}{b}$

$= > 1- \frac{p}{b} = 1 -cot \: a$

I hope it will helps you.... $\huge{\red{\ddot{\smile}}}$

Answer 2

$\displaystyle \sf{ \frac{1 - cosA}{sinA} } = tan \frac{A}{2}$

Given :

$\displaystyle \sf{ \frac{1 - cosA}{sinA} }$

To find :

The value of the expression

Formula Used :

$\displaystyle \sf{1. \: \: \: \: 1 - cosA = 2 {sin}^{2} \dfrac{A}{2} }$

$\displaystyle \sf{2. \: \: \: \: sinA = 2sin\dfrac{A}{2}cos\dfrac{A}{2}}$

Solution :

Step 1 of 2 :

Write down the given expression

Here the given expression is

$\displaystyle \sf{ \frac{1 - cosA}{sinA} }$

Step 2 of 2 :

Simplify the given expression

$\displaystyle \sf{ \frac{1 - cosA}{sinA} }$

$\displaystyle \sf{ = \frac{2 {sin}^{2} \dfrac{A}{2} }{2sin\dfrac{A}{2}cos\dfrac{A}{2}} }$

$\displaystyle \sf{ = \frac{ {sin}^{} \dfrac{A}{2} }{cos\dfrac{A}{2}} }$

$\displaystyle \sf{ = tan \dfrac{A}{2} }$

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