How do you prove that #costheta/(1-sintheta) + costheta/(1+sintheta) = 2 /costheta#?

1 Answer
Write your answer here...
Start with a one sentence answer
Then teach the underlying concepts
Don't copy without citing sources
preview
?

Answer

Write a one sentence answer...

Answer:

Explanation

Explain in detail...

Explanation:

I want someone to double check my answer

Describe your changes (optional) 200

1
Bdub Share
Mar 8, 2018

Answer:

See Below

Explanation:

#LHS : cos theta/(1-sin theta)+costheta/(1+sintheta)#

#=(costheta(1+sintheta)+costheta(1-sintheta))/((1-sintheta)(1+sin theta))#->common denominator

#=(costheta+sinthetacostheta+costheta-sinthetacostheta)/(1-sin^2theta)#

#=(costheta+cancel(sinthetacostheta)+costheta-cancel(sinthetacostheta))/(1-sin^2theta)#

#=(2costheta)/cos^2theta#->use the property #sin^2theta+cos^2theta=1#

#=(2cancelcostheta)/cos^cancel2theta#

#=2/costheta#

#=RHS#

Was this helpful? Let the contributor know!
1500
Trending questions
Impact of this question
38 views around the world
You can reuse this answer
Creative Commons License