SOLUTION: Sorry, but I don't understand how to do these sort of questions... Prove the following identities 2tan^2/1+tan^2 =2 sin^2 Sin^2(1+sec^2)=sec^2-cos^2 1-sinA+cosA/1-sinA= 1+sinA+

Question 836084: Sorry, but I don't understand how to do these sort of questions...
Prove the following identities
2tan^2/1+tan^2 =2 sin^2
Sin^2(1+sec^2)=sec^2-cos^2
1-sinA+cosA/1-sinA= 1+sinA+cosA/cosA

Please help, Thank you!
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Prove the following identities
2tan^2/1+tan^2 =2 sin^2
Note: 1+tan^2 = sec^2
----
So, 2tan^2/sec^2 = 2sin^2
Then 2[sin^2/cos^2]/[1/cos^2] = 2sin^2
Therefore: 2[sin^2/cos^2] = 2sin^2
Finally: 2sin^2 = 2sin^2
--------------------
Sin^2(1+sec^2)=sec^2-cos^2
---------
Multiply on the left side:
Convert sec^2 to tan^2+1 on the right side
sin^2 + tan^2 = tan^2+1 - cos^2
------
Rearrange the right side:
sin^2 + tan^2 = tan^2 + [1-cos^2]
------
Apply: sin^2 = 1-cos^2 to get
sin^2 + tan^2 = tan^2 + sin^2
========================================
1-sinA+cosA/1-sinA= 1+sinA+cosA/cosA
Comment: This is confusing.
Using parentheses, show where numerators and denominators begin and end.
==================
Cheers,
Stan

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