Verify the Identity 1/(1-sin(x))-1/(1+sin(x))=2tan(x)sec(x)

Problem

1/(1−sin(x))−1/(1+sin(x))=2*tan(x)*sec(x)

Solution

1. Find a common denominator for the fractions on the left side of the equation by multiplying the denominators (1−sin(x)) and (1+sin(x))

(1*(1+sin(x))−1*(1−sin(x)))/((1−sin(x))*(1+sin(x)))

2. Simplify the numerator by distributing the negative sign and combining like terms.

(1+sin(x)−1+sin(x))/((1−sin(x))*(1+sin(x)))

(2*sin(x))/((1−sin(x))*(1+sin(x)))

3. Expand the denominator using the difference of squares formula (a−b)*(a+b)=a2−b2

(2*sin(x))/(1−sin2(x))

4. Apply the Pythagorean identity cos2(x)+sin2(x)=1 which implies 1−sin2(x)=cos2(x) to rewrite the denominator.

(2*sin(x))/cos2(x)

5. Decompose the fraction into a product of two trigonometric ratios to match the right side of the identity.

2⋅sin(x)/cos(x)⋅1/cos(x)

6. Substitute the definitions tan(x)=sin(x)/cos(x) and sec(x)=1/cos(x) to reach the final form.

2*tan(x)*sec(x)

Final Answer

1/(1−sin(x))−1/(1+sin(x))=2*tan(x)*sec(x)

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