Respuesta :
[tex]\tan x+\cot x=\csc x.\sec x[/tex]
Let us change tan x and cot x to sin x and cos x
[tex]\begin{gathered} \because\tan x=\frac{\sin x}{\cos x} \\ \because\cot x=\frac{\cos x}{\sin x} \end{gathered}[/tex]Substitute them on the left side
[tex]LHS=\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}=\frac{\sin^2x+cox^2x}{\cos x\sin x}[/tex]I multiplied the denominators and multiply each numerator by the opposite denominator
[tex]\because\sin ^2x+cox^2x=1[/tex][tex]\therefore L.H.S=\frac{1}{\sin x\cos x}[/tex]Now we will work on the right hand side
[tex]\begin{gathered} \because\csc x=\frac{1}{\sin x} \\ \because\sec x=\frac{1}{\cos x} \end{gathered}[/tex]Substitute them on the right side
[tex]\because R.H.S=\frac{1}{\sin x}\times\frac{1}{\cos x}=\frac{1}{\sin x\cos x}[/tex]The L.H.S = R.H.S = 1/(sin x cos x)
