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For 
x such that 
0 < x < (pi)/(2), the expression 
(sqrt(1-cos^(2)x))/(sin x)+(sqrt(1-sin^(2)x))/(cos x) is equivalent to:
F. 0
G. 1
H. 2
J. 
-tan x
K. 
sin 2x

For x x such that 0<x<π2 0<x<\frac{\pi}{2} , the expression 1cos2xsinx+1sin2xcosx \frac{\sqrt{1-\cos ^{2} x}}{\sin x}+\frac{\sqrt{1-\sin ^{2} x}}{\cos x} is equivalent to:\newlineF. 00\newlineG. 11\newlineH. 22\newlineJ. tanx -\tan x \newlineK. sin2x \sin 2 x

Full solution

Q. For x x such that 0<x<π2 0<x<\frac{\pi}{2} , the expression 1cos2xsinx+1sin2xcosx \frac{\sqrt{1-\cos ^{2} x}}{\sin x}+\frac{\sqrt{1-\sin ^{2} x}}{\cos x} is equivalent to:\newlineF. 00\newlineG. 11\newlineH. 22\newlineJ. tanx -\tan x \newlineK. sin2x \sin 2 x
  1. Recognize Trigonometric Identities: Recognize that the expressions under the square roots are trigonometric identities. The identity sin2(x)+cos2(x)=1\sin^2(x) + \cos^2(x) = 1 can be rearranged to 1cos2(x)=sin2(x)1 - \cos^2(x) = \sin^2(x) and 1sin2(x)=cos2(x)1 - \sin^2(x) = \cos^2(x).
  2. Substitute Identities: Substitute the identities into the original expression to simplify it. This gives us (sin2(x))/(sinx)+(cos2(x))/(cosx)(\sqrt{\sin^2(x)})/(\sin x) + (\sqrt{\cos^2(x)})/(\cos x).
  3. Simplify Square Roots: Since we are given that 0<x<π20 < x < \frac{\pi}{2}, both sinx\sin x and cosx\cos x are positive in this range. Therefore, we can simplify sin2(x)\sqrt{\sin^2(x)} to sinx\sin x and cos2(x)\sqrt{\cos^2(x)} to cosx\cos x.
  4. Combine Terms: After simplification, the expression becomes sinxsinx+cosxcosx\frac{\sin x}{\sin x} + \frac{\cos x}{\cos x}, which simplifies to 1+11 + 1.
  5. Final Result: Adding the two terms together, we get 1+1=21 + 1 = 2.

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