Question

Tan theta /1-cot theta +cot theta /1-tan theta=sec theta ×cosec theta +1

Answer 1

tanФ/1-cotФ + cotФ/1-tanФ = secФcosecФ + 1

Taking LHS
   = tanФ/1-cotФ + cotФ/1-tanФ
Changing cotФ = 1/tanФ
  = tanФ/1-1/tanФ + cotФ/1-tanФ
  = tan²Ф/tanФ-1 + cotФ/1-tanФ
Changing Sign of 1-tanФ = -(tanФ-1)
  = tan²Ф/tanФ-1 - /tanФ-1
  = tan²Ф - cotФ/tanФ-1
Changing cotФ = 1/tanФ
  = tan²Ф-1/tanФ ÷ tanФ-1
  = tan³Ф-1/tanФ(tanФ-1)
As a³-b³=(a-b)(a²+b²+ab)
  = (tanФ-1)(tan²Ф+1+tanФ)/(tanФ-1)tanФ
Cancelling (tanФ-1) 
  = (tan²Ф+1+tanФ)/tanФ
As 1+tan²Ф=sec²Ф
  = sec²Ф+tanФ/tanФ
Also sec²Ф=1/cos²Ф   and tanФ=sinФ/cosФ
  = 1/cos²Ф ÷ sinФ/cosФ+tanФ/tanФ
  = 1/cos²Ф × cosФ/sinФ + 1
  = secФcosecФ+1  = RHS

:) Hope this Helps !!!

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:-

$\tt lhs \ratio -$

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

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

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

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

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

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

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

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

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

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

$= rhs$