Prove the following:
[1+sinx]/[1+cscx]=tanx/secx
=[1+sinx]/[1+1/sinx]
=[1+sinx]/[(sinx+1)/sinx]
=[1+sinx]*[sinx/(sinx+1)]
=[sinx+sin^2x]/[sinx+1]
=[sinx+(1-cos^2x)]/[sinx+1]
=???
This is where I'm stuck. Can someone help me. Please check what I got is right so far it's possible that I made a mistake somewhere.
Please and Thank you
LS = (1 + sinx)/(1 + 1/sinx)
= (1 + sinx)/( ( sinx + 1)/sinx )
= (1 + sinx) ( sinx/(1+sinx))
= sinx
RS = (sinx/cosx)(cosx)
= sinx
= LS
so I have to do both sides but I thought can only do one side
2x
To continue simplifying the expression, you can use the identity sin^2(x) + cos^2(x) = 1.
=[sinx+(1-cos^2x)]/[sinx+1]
= [sinx + 1 - cos^2x]/[sinx + 1]
Now, we can rewrite cos^2x as 1 - sin^2x using the identity stated above.
= [sinx + 1 - (1 - sin^2x)]/[sinx + 1]
= [sinx + 1 - 1 + sin^2x]/[sinx + 1]
= [sin^2x + sinx]/[sinx + 1]
Next, we can factor out sinx from the numerator.
= sinx * (sinx + 1)/[sinx + 1]
Finally, the sinx + 1 terms in the numerator and denominator cancel each other out.
= sinx
Therefore, [1+sinx]/[1+cscx] simplifies to sinx.