Question

(cotx+cscx)/(sinx+tanx)

Answer 1

Answer:
`\bold{(cot(x))/(sin(x))}`

Explanation:

Convert everything to "sin" and "cos" and then cancel out the common factors.

`(cot(x)+csc(x))/(sin(x)+tan(x))\\\\\\\bigg((cos(x))/(sin(x))+(1)/(sin(x))\bigg)/\bigg((sin(x))/(1)+(sin(x))/(cos(x))\bigg)\\\\\\\bigg((cos(x))/(sin(x))+(1)/(sin(x))\bigg)/\bigg[(sin(x))/(1)\bigg((cos(x))/(cos(x))\bigg)+(sin(x))/(cos(x))\bigg]\\\\\\\bigg((cos(x))/(sin(x))+(1)/(sin(x))\bigg)/\bigg((sin(x)cos(x))/(cos(x))+(sin(x))/(cos(x))\bigg)`

`\text{Simplify:}\\\\\bigg((cos(x)+1)/(sin(x))\bigg)/\bigg((sin(x)cos(x)+sin(x))/(cos(x))\bigg)\\\\\\\text{Multiply by the reciprocal (fraction rules)}:\\\\\bigg((cos(x)+1)/(sin(x))\bigg)*\bigg((cos(x))/(sin(x)cos(x)+sin(x))\bigg)\\\\\\\text{Factor out the common term on the right side denominator}:\\\\\bigg((cos(x)+1)/(sin(x))\bigg)*\bigg((cos(x))/(sin(x)(cos(x)+1))\bigg)`

`\text{Cross out the common factor of (cos(x) + 1) from the top and bottom}:\\\\\bigg((1)/(sin(x))\bigg)*\bigg((cos(x))/(sin(x))\bigg)\\\\\\\bigg((1)/(sin(x))\bigg)* cot(x)}\qquad \rightarrow \qquad (cot(x))/(sin(x))`