Math, asked by ankitroy9744, 1 year ago

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

Answers

Answered by Outsss
1

Answer:

I have the solution

See the attachment

Thanks for asking.

Attachments:
Answered by dhyan531
0

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

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