Is (1+sec(x))/(csc(x))-tan(x)=sin(x) ?

The answer to whether (1+sec(x))/(csc(x))-tan(x)=sin(x) is True

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prove (1+sec(x))/(csc(x))-tan(x)=sin(x)

prove

\frac{1 + sec ( x )}{csc ( x )} - tan (x) = sin (x)

True

Solution steps

\frac{1 + sec ( x )}{csc ( x )} - tan (x) = sin (x)

Manipulating left side

\frac{1 + sec ( x )}{csc ( x )} - tan (x)

Express with sin, cos

\frac{1 + sec ( x )}{csc ( x )} - tan (x)

Use the basic trigonometric identity:

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

= \frac{1 + \frac{1}{c o s ( x )}}{csc ( x )} - tan (x)

Use the basic trigonometric identity:

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

= \frac{1 + \frac{1}{c o s ( x )}}{\frac{1}{s i n ( x )}} - tan (x)

Use the basic trigonometric identity:

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

= \frac{1 + \frac{1}{c o s ( x )}}{\frac{1}{s i n ( x )}} - \frac{sin ( x )}{cos ( x )}

Simplify

\frac{1 + \frac{1}{c o s ( x )}}{\frac{1}{s i n ( x )}} - \frac{sin ( x )}{cos ( x )} : sin (x)

\frac{1 + \frac{1}{c o s ( x )}}{\frac{1}{s i n ( x )}} - \frac{sin ( x )}{cos ( x )}

\frac{1 + \frac{1}{c o s ( x )}}{\frac{1}{s i n ( x )}} = \frac{sin ( x ) ( cos ( x ) + 1 )}{cos ( x )}

\frac{1 + \frac{1}{c o s ( x )}}{\frac{1}{s i n ( x )}}

Apply the fraction rule:

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

= \frac{( 1 + \frac{1}{c o s ( x )} ) sin ( x )}{1}

Join

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

Multiply:

1 \cdot cos (x) = cos (x)

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

= \frac{\frac{c o s ( x ) + 1}{c o s ( x )} sin ( x )}{1}

Apply the fraction rule:

\frac{a}{1} = a

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

Multiply fractions:

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

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

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

Apply rule

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

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

Factor

(cos (x) + 1) sin (x) - sin (x) : sin (x) cos (x)

(cos (x) + 1) sin (x) - sin (x)

Factor out common term

sin (x)

= sin (x) (1 + cos (x) - 1)

Refine

= sin (x) cos (x)

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

Cancel the common factor:

cos (x)

= sin (x)

= sin (x)

= sin (x)

We showed that the two sides could take the same form

\Rightarrow True

2 sin (x) = \frac{4 cos ( x ) - 1}{tan ( x )}

cos (8 x) = cos^{2} (4 x) - sin^{2} (4 x)

tan (\frac{π}{2} - x) cot (x) = csc^{2} (x) - 1

cot^{2} (θ) cos (2 θ) = cot^{2} (θ) - 2 cos^{2} (θ)

\frac{cot ^{2} ( x )}{cos ( x )} = cot (x) \cdot csc (x)

Is (1+sec(x))/(csc(x))-tan(x)=sin(x) ?
The answer to whether (1+sec(x))/(csc(x))-tan(x)=sin(x) is True