Question

Prove it :


`(1 + \cos(x) - \sin(x) )/(1 + \cos(x) + \sin(x) ) = \sec(x) - \tan(x)`​
Please show working using paper...

Answer 1

Heya!

`\text{First, manipulate the left side.}\\\text{Use the rule:}~a-b=((a-b)(a-b))/(a+b) = (a^2-b^2)/(a+b) \\\text{Use the identity:}~1-sin^2(x)=cos^2(x)\\(1+cos(x)-sin(x))/(1+cos(x)+sin(x)) = (cos(x)+(cos^2(x))/(1+sin(y)))/(1+cos(x)+sin(x))`

`\text{Second, simplify the numerator}~cos(x)+(cos^2(x))/(1+sin(x))\\\text{Convert element to a fraction:}~cos(x)=(cos(x)(1+sin(x)))/(1+sin(x))\\\text{Add:}~cos(x)=(cos(x)(1+sin(x)))/(1+sin(x))+(cos^2(x))/(1+sin(x))\\\text{The denominators are equal so combine:}~(cos(x)(1+sin(x))+cos^2(x))/(1+sin(x))\\\text{Simplify:} ((cos(x)(sin(x)+1)+cos^2(x))/(1+sin(x)) )/(1+cos(x)+sin(x))\\ \text{Apply fraction rule:}~(cos(x)(1+sin(x))+cos^2(x))/(1+sin(x))\\\\`

`\text{Factor:}~(cos(x)(1+sin(x)+cos(x)))/((1+sin(x))(1+cos(x)+sin(x))) \\\text{Simplify:}~(cos(x))/(1+sin(x))`

`\text{Third, manipulate the right side.}\\\text{Use the basic trigonometric identity:}~sec(x)=(1)/(cos(x)) \\\text{Use the basic trigonometric identity:}~tan(x)=(sin(x))/(cos(x)) \\\text{Put the expression back together:}~(1)/(cos(x))-(sin(x))/(cos(x))\\\text{Simplify:}~((cos^2(x))/(1+sin(x)) )/(cos(x))`

`\text{Fourth, simplify.}\\\text{Apply the fraction rule:}~(cos^2(x))/((1+sin(x))cos(x)) \\\text{Cancel out the common factor:}~(cos(x))/(1+sin(x))`

Therefore, the expression is TRUE.

Best of Luck!