Posted by **mtd** on Wednesday, April 9, 2008 at 1:37pm.

square root(a^2 -u^2) where a>0, let u=(a sinx) where -pi/2<x<pi/2

answer is (a cosx) and I don't know how to get there

## Answer this Question

## Related Questions

- Trig........ - I need to prove that the following is true. Thanks (cosx / 1-sinx...
- trigonometry - can i use factoring to simplify this trig identity? the problem ...
- Trigonometry Check - Simplify #3: [cosx-sin(90-x)sinx]/[cosx-cos(180-x)tanx] = [...
- Trigonometry. - ( tanx/1-cotx )+ (cotx/1-tanx)= (1+secxcscx) Good one! ...
- Math - Verify the identity . (cscX-cotX)^2=1-cosX/1+cosX _______ sorry i cant ...
- Math - 1) evaluate without a calculator: a)sin(3.14/4) b) cos(-3(3.14)/4) c) ...
- maths - trigonometry - I've asked about this same question before, and someone ...
- Pre-Calc - Trigonometric Identities Prove: (tanx + secx -1)/(tanx - secx + 1)= ...
- Trigonometry - Simplify #1: cscx(sin^2x+cos^2xtanx)/sinx+cosx = cscx((1)tanx)/...
- Trigonometry - Prove the following trigonometric identities. please give a ...