Question

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Answer 1

Question :-

Prove that :

$\rm{ \frac{ \tan \theta}{1 - \cot \theta} + \frac{ \cot \theta }{1 - \tan \theta} = 1 + \sec \theta \cosec \theta}$

Solution :-

Here taking LHS.

We know that,

$\rm:\longmapsto{ \cot \theta = \frac{1}{ \tan \theta } \: or \: \frac{ \cos \theta }{ \sin \theta } }$

on using this,

$\rm:\longmapsto{ \frac{ \tan \theta }{1 - \frac{1}{ \tan \theta} } + \frac{ \frac{1}{ \tan \theta } }{1 - \tan \theta } } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\\rm:\longmapsto{ \frac{ \tan \theta }{ \frac{ \tan \theta - 1 }{ \tan \theta} } + \frac{1}{ \tan \theta(1 - \tan \theta )} } \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\\rm:\longmapsto{ \frac{ \tan {}^{2} \theta}{ \tan \theta - 1} + \frac{1}{ \tan \theta(1 - \tan \theta ) } } \: \: \: \: \: \: \: \: \\ \\ \rm:\longmapsto{ \frac{ \tan {}^{2} \theta }{ \tan \theta - 1 } - \frac{1 }{ \tan \theta(1 - \tan \theta ) } } \: \: \: \: \: \: \: \\ \\ \rm:\longmapsto{ \frac{ \tan {}^{3} \theta - 1 }{ \tan \theta(1 - \tan \theta ) } } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \rm:\longmapsto{ \frac{ \cancel{(\tan \theta - 1)}( \tan {}^{2} \theta + \tan \theta + 1)}{ \tan \theta \cancel{( \tan \theta - 1 ) }} } \\ \\ \rm:\longmapsto{ \frac{ \tan {}^{2} \theta + \tan \theta + 1 }{ \tan \theta} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

Also we can write

$\rm:\longmapsto{ \frac{ \tan {}^{2} \theta }{ \tan \theta } + \cancel\frac{ \tan \theta}{ \tan \theta} + \frac{1}{ \tan \theta} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:\\ \\ \rm:\longmapsto{ \tan \theta + 1 + \cot \theta \: \: \: \bigg \{ \because \frac{1}{ \tan \theta} = \cot \theta \bigg \}}$

We know that,

$\rm:\longmapsto{ \tan \theta = \frac{ \sin \theta}{ \cos \theta} } \\ \\ \rm:\longmapsto{ \cot \theta = \frac{ \cos \theta}{ \sin \theta} }$

On using this,

$\rm:\longmapsto{1 + \frac{ \sin \theta}{ \cos \theta} + \frac{ \cos \theta}{ \sin \theta} } \: \: \: \\ \\ \rm:\longmapsto{1 + \frac{ \sin {}^{2} \theta + \cos {}^{2} \theta }{ \sin \theta \cos \theta} }$

We know that,

$\rm:\longmapsto{ \sin {}^{2} \theta + \cos {}^{2} \theta = 1 }$

On using this,

$\rm:\longmapsto{1 + \frac{1}{ \sin \theta \cos \theta} }$

Also we can write,

$\rm:\longmapsto{1 + \frac{1}{ \sin \theta } \times \frac{1}{ \cos \theta} }$

We know that,

$\rm:\longmapsto{ \frac{1}{ \sin \theta} = \cosec \theta } \\ \\ \rm:\longmapsto{ \frac{1}{ \cos \theta} = \sec \theta } \: \:$

Hence,

$\rm:\longmapsto{ 1 + \sec \theta \cosec \theta}$

LHS = RHS

• Hence Proved!

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