Login or Sign Up
Menu
Ask a New Question

I need to prove that the following is true. Thanks.

csc^2(A/2)=2secA/secA-1

Right Side=(2/cosA)/(1/cosA - 1)
= (2/cosA)/[(1-cosA)/cosA]
=2/cosA x (cosA)/(1-cosA)
=2/(1-cosA)

now recall cos 2X = cos^2 X - sin^2 X and we could say
cos A = cos^2 A/2 - sin^2 A/2 and of course sin^2 A/2 + cos^2 A/2=1

so the above 2/(1 - cosA)
=2/[sin^2 A/2 + cos^2 A/2 - (cos^2 A/2 - sin^2 A/2)]
= 2/(2sin^2 A/2)
= csc^2 A/2
= Left Side

Ok, so just using a different form of the double/half-angle formula allowed you to get the sin and therein getting csc.

Thanks Reiny! For a detailed how to and detailed steps. I will be using this service in the future. Thanks again.

You're welcome! I'm glad I could help. Just remember, when it comes to math proofs, sometimes thinking outside the box can lead to some interesting results. And if you ever need any more assistance in the future, feel free to reach out. Happy math-ing!

You're welcome! I'm glad I could help. If you have any more questions or need assistance with anything else in the future, feel free to ask. I'm here to help you. Have a great day!

You're welcome! I'm glad I could help you understand the proof. Using different forms of trigonometric identities and manipulating them is often the key to solving trigonometric equations. If you have any more questions or need further explanations in the future, feel free to ask. Thanks for using this service, and have a great day!