Question

Prove that tan theta/1-cot theta +cot theta/1-tan theta - 1 = sec theta * cosec theta

Answer 1

Answer:

Proved

Step-by-step explanation:

let theta = ∝

tan theta/1-cot theta +cot theta/1-tan theta - 1 = sec theta * cosec theta

tan ∝/1-cot ∝ +cot ∝/1-tan ∝ - 1 = sec ∝* cosec∝

LHS =  tan ∝/1-cot ∝ +cot ∝/1-tan ∝ - 1 = tan ∝/1-1/tan∝ +1/tan ∝/1-tan ∝ - 1

= tan² ∝/tan∝-1 +1/tan∝(1-tan ∝) - 1 = tan² ∝/tan∝-1 -1/tan∝(tan ∝-1) - 1

= tan³∝ - 1/tan∝(tan ∝-1)  - 1 = tan²∝ +tan∝ +1 / tan∝ - 1 = tan∝ + cot∝ + 1-1

=  tan∝ + cot∝ = sin∝/cos∝ + cos∝/sin∝ =  sin²∝ + cos²∝/sin∝ cos∝

= 1/sin∝ cos∝ = sec ∝* cosec∝ = RHS

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:-

$\tt lhs \ratio -$

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

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

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

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

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

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

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

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

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

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

$= rhs$