Question

cos B /1- tan B + sin B / 1- cot B = sin B + cos B prove that​

Answer 1

Solution →

$\frac{cos \: b}{1 - \: tan \: b} \: + \: \frac{sin \: b}{1 - \: cot \: b} = \: sin \: b \: + \: cos \: b$

Let's Take LHS →

$\frac{cos \: b}{1 - \: tan \: b} \: + \: \frac{sin \: b}{1 - \: cot \: b}$

$\frac{cos \: b}{1 - \frac{sin \: b}{cos \: b} } \: + \: \frac{sin \: b}{1 - \frac{cos \: b}{sin \: b} }$

$\frac{cos \: b}{ \frac{cos \: b \: - \: sin \: b}{cos \: b} } + \frac{sin \: b}{ \frac{sin \: b \: - cos \: b}{sin \: b} }$

$\frac{ {cos}^{2}{b} }{cos \: b \: - \: sin \: b} + \frac{ {sin}^{2}{b} }{sin \: b \: - cos \: b}$

$\frac{ {cos}^{2}{b} }{cos \: b \: - \: sin \: b} + ( - \frac{ {sin}^{2}{b} }{cos \: b \: - sin \: \: b} )$

$\frac{ {cos}^{2}{b} }{cos \: b \: - \: sin \: b} - \frac{ {sin}^{2}{b} }{cos \: b \: - \: sin \: b}$

$\frac{ {cos}^{2}b - {sin}^{2}b }{cos \: b \: - \: sin \: b}$

Now (cos²b - sin²b ) is like (a²-b²),

So the formula of (a²-b²) is

→ (a+b)(a-b)

$\frac{(cos \: b \: + \: sin \: b)(cos \: b \: - \: sin \: b)}{(cos \: b - sin \: b \: )}$

Now (cos b - sin b ) will be cancelled.

$(cos \: b \: </strong><strong>+</strong><strong> </strong><strong> \: sin \: b)$

Or we can write it like ,

$</strong><strong>\</strong><strong>b</strong><strong>o</strong><strong>x</strong><strong>e</strong><strong>d</strong><strong>{</strong><strong>(</strong><strong>Sin</strong><strong> \: b \: + \: </strong><strong>Cos</strong><strong> \: b)</strong><strong>}</strong><strong>$ RHS

Hence Proved !!

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