I need help in both of these identities. I have been stuck in these two forever
#need urgent help in trigonometric identities
tight ocean
left shoal
Don't invade others' threads. Create your own.
left shoal
I need help in both of these identities. I have been stuck in these two forever
(y)
Use the difference of squares formula for the right side.
(z)
cos(A)/(csc(A) ± 1) = sin(A)cos(A)/(1 ± sin(A))
So:
2sin(A)/cos(A) = sin(A)cos(A)/(1 - sin(A)) + sin(A)cos(A)/(1 + sin(A))
Divide both sides by sin(A)cos(A).
2/cos(A)^2 = 1/(1 - sin(A)) + 1/(1 + sin(A))
Now just work with the right side.
tight ocean
I dont get number y
I thought we need to solve the left hand side to prove the right hand side
Or is it possible to solve the right hand side in order to prove with the left hand side?
@left shoal
lunar lichen
I thought we need to solve the left hand side to prove the right hand side
a^2 -b^2 = (a+b)(a-b)
(cos^2x)^2 - (sin^2x)^2 = (cos^2x - sin^2x)(cos^2x+sin^2x)
Notice anything special about either terms?
lunar lichen
And anything times 1 is itself
tight ocean
Yes i know
But my question was about whether solving which side would matter or not
In this case the right hand side is being solved
However in the book all example sums are solved using the left hand side so i assumed thats the rule
Does the side matter?
lunar lichen
No, as long as you reach the final conclusion
tight ocean
No, as long as you reach the final conclusion
Okay thanks
lunar lichen
Idk why your book does that, likely concidence
tight ocean
That clears up my confusion
fair riverBOT
@glass comet has given 1 rep to @lunar lichen
lunar lichen
Or easier to write proofs on one side only
lunar lichen
You could theoretically solve on left side by noticing the multiplication by sin^2x + cos^2x but that’s not a very obvious jump
tight ocean
Yup