Question

prove that : tan theta /1-cot theta +cot theta /1-tan theta =1+sec theta cosec theta.

Answer 1

HELLO DEAR,



GIVEN:- tanθ/(1 - cotθ) + cotθ/(1 - tanθ)

=> tanθ/(1 - 1/tanθ) + (1/tanθ)/(1 - tanθ)

=> tan²θ/(tanθ - 1) - 1/tanθ(tanθ - 1)

=> 1/(tanθ - 1) { tan²θ - 1/tanθ }

=> 1/(tanθ - 1) { (tan³θ - 1)/tanθ)

[as, a³ - b³ = (a - b)(a² + b² + ab)

=> {(tanθ - 1)(tan²θ + 1 + tanθ)}/{(tanθ - 1)(tanθ)}

=> tanθ + cotθ + 1

=> sinθ/cosθ + cosθ/sinθ + 1

=> (sin²θ + cos²θ)/sinθ . cosθ + 1

=> 1/sinθ . cosθ + 1

=> cosecθ . secθ + 1



I HOPE IT'S HELP YOU DEAR,
THANKS

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:-

$\tt lhs \ratio -$

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

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

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

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

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

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

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

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

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

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

$= rhs$