# Verify the trigonometric identity square root [(1+sin x)/(1-sinx)]=(1+sinx)/|cosx|

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### 2 Answers

We have to verify: sqrt [(1 + sin x)/(1 - sin x)] = (1 + sin x)/|cos x|

sqrt [(1 + sin x)/(1 - sin x)]

=> sqrt [(1 + sin x)(1 + sin x)/(1 - sin x)(1 + sin x)]

=> sqrt [(1 + sin x)^2/(1 - (sin x)^2)]

=> sqrt [(1 + sin x)^2/(cos x)^2]

=> (1 + sin x)/|cos x|

**This verifies that sqrt [(1 + sin x)/(1 - sin x)] = (1 + sin x)/|cos x|**

We'll manage the left side of the expression and we'll multiply the numerator and denominator by (1+sin x).

sqrt [(1+sin x)^2/(1-sin x)*(1+sin x)] = (1+sinx)/|cosx|

We'll recognize the difference of 2 squares at denominator:

sqrt [(1+sin x)^2/[1- (sin x)^2]

We'll apply the Pythagorean identity:

1- (sin x)^2 = (cos x)^2

sqrt [(1+sin x)^2/[1- (sin x)^2] = |(1+sin x)^2|/|cos x|

Since 1 + sin x >= 0 => |(1+sin x)| = (1+sin x)

sqrt [(1+sin x)^2/[1- (sin x)^2] = (1+sin x)/|cos x|

**We notice that LHS = RHS, therefore the identity sqrt [(1+sin x)^2/[1- (sin x)^2] = (1+sin x)/|cos x| is verified.**