Question

Cosx(tanx+cotx)=

Pls Help

Answer 1

→cosecx

Explanation:

`\hookrightarrow cosx(tanx+cotx)\\\\\hookrightarrow cosx((sinx)/(cosx) +(cosx)/(sinx) )\\\\\hookrightarrow cosx((sinx * sinx + cosx * cosx)/(cosx*sinx) )\\\\\hookrightarrow cosx((sin^(2)x+cosx^(2)x)/(cosx*sinx) )\\\\\hookrightarrow cosx((1)/(cosx*sinx) )\\\\\hookrightarrow cosx * (1)/(cosx) * (1)/(sinx) \\\\\hookrightarrow 1*(1)/(sinx) \\\\\hookrightarrow (1)/(sinx) \\\\\hookrightarrow cosecx`

Answer 2

Answer is csc(x)

Explanation:

I convert tan(x)= sin(x)/cos(x) and cot(x)= cos(x)/sin(x).

Then distribute the cos(x).

Then turn it into fractor that has same denominator.

Simplified the numerator sin^2(x)+cos^2(x)=1

`cos(x)*\((tan(x)+cot(x))\\\\cos(x)*((sin(x))/(cos(x)) +(cos(x))/(sin(x)))\\\\sin(x)+(cos^2(x))/(sin(x))\\\\sin(x)*(sin(x))/(sin(x))+(cos^2(x))/(sin(x))\\\\(sin^2(x))/(sin(x))+(cos^2(x))/(sin(x))\\\\(sin^2(x)+cos^2(x))/(sin(x))\\\\(1)/(sin(x))\\\\csc(x)`