Is csc(x)*tan(x)+sec(x)=2sec(x) ?

The answer to whether csc(x)*tan(x)+sec(x)=2sec(x) is True

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prove csc(x)*tan(x)+sec(x)=2sec(x)

prove

csc (x) \cdot tan (x) + sec (x) = 2 sec (x)

True

Solution steps

csc (x) tan (x) + sec (x) = 2 sec (x)

Manipulating left side

csc (x) tan (x) + sec (x)

Express with sin, cos

sec (x) + csc (x) tan (x)

Use the basic trigonometric identity:

sec (x) = \frac{1}{cos ( x )}

= \frac{1}{cos ( x )} + csc (x) tan (x)

Use the basic trigonometric identity:

csc (x) = \frac{1}{sin ( x )}

= \frac{1}{cos ( x )} + \frac{1}{sin ( x )} tan (x)

Use the basic trigonometric identity:

tan (x) = \frac{sin ( x )}{cos ( x )}

= \frac{1}{cos ( x )} + \frac{1}{sin ( x )} \cdot \frac{sin ( x )}{cos ( x )}

Simplify

\frac{1}{cos ( x )} + \frac{1}{sin ( x )} \cdot \frac{sin ( x )}{cos ( x )} : \frac{2}{cos ( x )}

\frac{1}{cos ( x )} + \frac{1}{sin ( x )} \cdot \frac{sin ( x )}{cos ( x )}

\frac{1}{sin ( x )} \cdot \frac{sin ( x )}{cos ( x )} = \frac{1}{cos ( x )}

\frac{1}{sin ( x )} \cdot \frac{sin ( x )}{cos ( x )}

Multiply fractions:

\frac{a}{b} \cdot \frac{c}{d} = \frac{a \cdot c}{b \cdot d}

= \frac{1 \cdot sin ( x )}{sin ( x ) cos ( x )}

Cancel the common factor:

sin (x)

= \frac{1}{cos ( x )}

= \frac{1}{cos ( x )} + \frac{1}{cos ( x )}

Apply rule

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{1 + 1}{cos ( x )}

Add the numbers:

1 + 1 = 2

= \frac{2}{cos ( x )}

= \frac{2}{cos ( x )}

= \frac{2}{cos ( x )}

Rewrite using trig identities

Use the basic trigonometric identity:

cos (x) = \frac{1}{sec ( x )}

\frac{2}{\frac{1}{s e c ( x )}}

Simplify

\frac{2}{\frac{1}{s e c ( x )}}

Apply the fraction rule:

\frac{a}{\frac{b}{c}} = \frac{a \cdot c}{b}

= \frac{2 sec ( x )}{1}

Apply rule

\frac{a}{1} = a

= 2 sec (x)

2 sec (x)

2 sec (x)

We showed that the two sides could take the same form

\Rightarrow True

Is csc(x)*tan(x)+sec(x)=2sec(x) ?
The answer to whether csc(x)*tan(x)+sec(x)=2sec(x) is True