Prove that tan theta /1-cot theta+cot theta/1-tan theta=1+sec theta.Cosec theta
Answers
Answer:
I have the solution
See the attachment
Thanks for asking.
GIVEN:-
\sf \frac{ \tan \theta}{1 - cot\theta } + \frac{ \cot \theta}{ 1 - \tan \theta} = 1 + \sec \theta \cosec \theta
1−cotθ
tanθ
+
1−tanθ
cotθ
=1+secθcosecθ
TO FIND:-
\tt lhs = rhslhs=rhs
SOLUTION:-
\tt lhs \ratio -lhs:−
\sf \frac{ \tan \theta}{1 - cot\theta } + \frac{ \cot \theta}{ 1 - \tan \theta}
1−cotθ
tanθ
+
1−tanθ
cotθ
\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 }}=
1−
sinθ
cosθ
cosθ
sinθ
+
1−
cosθ
sinθ
sinθ
cosθ
\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 }}=
sinθ
sinθ−cosθ
cosθ
sinθ
+
cosθ
cosθ−sinθ
sinθ
cosθ
\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 ) }=
cosθ
sinθ
×
(sinθ−cosθ)
sinθ
+
sinθ
cosθ
×
(cosθ−sinθ)
cosθ
\tt = \frac{ { \sin}^{2}\theta}{ \cos \theta( \sin \theta - \cos \theta) } - \frac{ { \cos }^{2} \theta }{ \sin \theta( \sin \theta - \cos \theta ) }=
cosθ(sinθ−cosθ)
sin
2
θ
−
sinθ(sinθ−cosθ)
cos
2
θ
\tt = \frac{1}{( \sin \theta - \cos \theta) } [ \frac{ { \sin}^{2} \theta }{ \cos \theta } - \frac{ { \cos }^{2} \theta}{ \sin \theta } ]=
(sinθ−cosθ)
1
[
cosθ
sin
2
θ
−
sinθ
cos
2
θ
]
\tt = \frac{1}{( \sin \theta - \cos \theta) } [ \frac{ { \sin}^{3 } \theta - { \cos }^{3} \theta }{ \sin \theta \cos \theta } ]=
(sinθ−cosθ)
1
[
sinθcosθ
sin
3
θ−cos
3
θ
]
\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 } ]=
(sinθ−cosθ)
1
[
sinθcosθ
(sinθ−cosθ)(sin
2
θ+cos
2
θ+sinθcosθ)
]
\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) }=
(sinθcosθ)
(1+sinθcosθ)
=
sinθcosθ
1
+
(sinθcosθ)
(sinθcosθ)
= \sec \theta \cosec \theta + 1=secθcosecθ+1
= rhs=rhs