Question

Prove tan theta /1-cot theta +cot theta /1-tan theta = 1+sec theta . Cosec theta

Answer 1

LHS:

$\frac{ \tan(a) }{1 - \cot(a) } + \frac{ \cot(a) }{1 - \tan(a) }$

$\frac{ \tan(a) }{1 - \frac{1}{ \tan(a) } } - \frac{ \cot(a) }{ \tan(a) - 1} \:$

$\frac{ \tan(a) }{ \frac{ \tan( a) - 1 }{ \tan(a) } } - \frac{ \cot(a) }{ \tan(a) - 1}$

$\frac{ \tan {}^{2} (a) }{ \tan(a) - 1 } - \frac{ \cot(a) }{ \tan(a) - 1 }$

$\frac{ \tan {}^{2} (a) - \cot(a) }{ \tan(a) - 1 }$

HOPE YOU UNDERSTOOD THE STEPS!!!

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Answer 2

GIVEN:-

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

TO FIND:-

$\tt lhs = rhs$

SOLUTION:-

$\tt lhs \ratio -$

$\sf \frac{ \tan \theta}{1 - cot\theta } + \frac{ \cot \theta}{ 1 - \tan \theta}$

$\tt = \frac{ \frac{ \sin \theta }{ \cos\theta } }{1 - \frac{\cos\theta}{\sin \theta} } + \frac{\frac{\cos\theta}{\sin \theta} }{1 - \frac{ \sin \theta }{ \cos\theta }}$

$\tt = \frac{ \frac{ \sin \theta }{ \cos\theta } }{ \frac{ \sin \theta- \cos\theta}{ \sin \theta} } + \frac{\frac{\cos\theta}{\sin \theta} }{ \frac{ \cos\theta - \sin \theta }{ \cos\theta }}$

$\tt = \frac{ \sin \theta }{ \cos \theta } \times \frac{ \sin \theta }{(\sin \theta - \cos \theta ) } + \frac{\cos \theta }{\sin \theta} \times \frac{ \cos \theta }{( \cos \theta - \sin \theta ) }$

$\tt = \frac{ { \sin}^{2}\theta}{ \cos \theta( \sin \theta - \cos \theta) } - \frac{ { \cos }^{2} \theta }{ \sin \theta( \sin \theta - \cos \theta ) }$

$\tt = \frac{1}{( \sin \theta - \cos \theta) } [ \frac{ { \sin}^{2} \theta }{ \cos \theta } - \frac{ { \cos }^{2} \theta}{ \sin \theta } ]$

$\tt = \frac{1}{( \sin \theta - \cos \theta) } [ \frac{ { \sin}^{3 } \theta - { \cos }^{3} \theta }{ \sin \theta \cos \theta } ]$

$\tt = \frac{1}{( \sin \theta - \cos \theta) } [ \frac{( \sin \theta - \cos \theta)( { \sin}^{2} \theta + { \cos }^{2} \theta + \sin \theta \cos \theta )}{ \sin \theta \cos \theta } ]$

$\tt = \frac{(1 + \sin \theta \cos \theta)}{( \sin \theta \cos \theta)} = \frac{1}{ \sin \theta \cos \theta } + \frac{( \sin \theta \cos \theta) }{( \sin \theta \cos \theta) }$

$= \sec \theta \cosec \theta + 1$

$= rhs$