show that trigonometric identity!!!!!!!!!!!!!!

Question

tan(x)+sec(x)-1/tan(x)+sec(x)-1=1+sin(x)/cos(x)

Answer 1

LHS = (tan(x)+sec(x)-1) / (tan(x)-sec(x)+1)

Apply Pythagorean formula : sec²(x) - tan²(x) = 1

= [tan(x)+sec(x) -(sec²(x) - tan²(x))] / (tan(x)-sec(x)+1)

formula : a²- b² = (a + b)(a - b)

= [tan(x)+sec(x) - {(tan(x)+sec(x))(sec(x) - tan(x))}] /  (tan(x)-sec(x)+1)

Take out common factors.

= [tan(x)+sec(x){1 - (sec(x) - tan(x))}] / (tan(x)-sec(x)+1)

= [tan(x)+sec(x){(tan(x)-sec(x)+1)}] / (tan(x)-sec(x)+1)

Cancel common terms.

= tan(x)+sec(x)

= [sin(x)/cos(x)] +[1/cos(x)]

= [1 + sin(x)] /cos(x)

RHS = [1 + sin(x)] /cos(x)