Question

Show tan (x) + cot (x) • sec(x)

Answer 1

$\tan x+\cot x=\csc x.\sec x$

Let us change tan x and cot x to sin x and cos x

$\begin{gathered} \because\tan x=\frac{\sin x}{\cos x} \\ \because\cot x=\frac{\cos x}{\sin x} \end{gathered}$

Substitute them on the left side

$LHS=\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}=\frac{\sin^2x+cox^2x}{\cos x\sin x}$

I multiplied the denominators and multiply each numerator by the opposite denominator

$\because\sin ^2x+cox^2x=1$$\therefore L.H.S=\frac{1}{\sin x\cos x}$

Now we will work on the right hand side

$\begin{gathered} \because\csc x=\frac{1}{\sin x} \\ \because\sec x=\frac{1}{\cos x} \end{gathered}$

Substitute them on the right side

$\because R.H.S=\frac{1}{\sin x}\times\frac{1}{\cos x}=\frac{1}{\sin x\cos x}$

The L.H.S = R.H.S = 1/(sin x cos x)