Is (2+csc(x))/(sec(x))-2cos(x)=cot(x) ?

The answer to whether (2+csc(x))/(sec(x))-2cos(x)=cot(x) is True

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prove (2+csc(x))/(sec(x))-2cos(x)=cot(x)

prove

\frac{2 + csc ( x )}{sec ( x )} - 2 cos (x) = cot (x)

True

Solution steps

\frac{2 + csc ( x )}{sec ( x )} - 2 cos (x) = cot (x)

Manipulating left side

\frac{2 + csc ( x )}{sec ( x )} - 2 cos (x)

Express with sin, cos

\frac{2 + csc ( x )}{sec ( x )} - 2 cos (x)

Use the basic trigonometric identity:

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

= \frac{2 + \frac{1}{s i n ( x )}}{sec ( x )} - 2 cos (x)

Use the basic trigonometric identity:

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

= \frac{2 + \frac{1}{s i n ( x )}}{\frac{1}{c o s ( x )}} - 2 cos (x)

Simplify

\frac{2 + \frac{1}{s i n ( x )}}{\frac{1}{c o s ( x )}} - 2 cos (x) : \frac{cos ( x )}{sin ( x )}

\frac{2 + \frac{1}{s i n ( x )}}{\frac{1}{c o s ( x )}} - 2 cos (x)

\frac{2 + \frac{1}{s i n ( x )}}{\frac{1}{c o s ( x )}} = \frac{cos ( x ) ( 2 sin ( x ) + 1 )}{sin ( x )}

\frac{2 + \frac{1}{s i n ( x )}}{\frac{1}{c o s ( x )}}

Apply the fraction rule:

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

= \frac{( 2 + \frac{1}{s i n ( x )} ) cos ( x )}{1}

Join

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

2 + \frac{1}{sin ( x )}

Convert element to fraction:

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

= \frac{2 sin ( x )}{sin ( x )} + \frac{1}{sin ( x )}

Since the denominators are equal, combine the fractions:

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

= \frac{2 sin ( x ) + 1}{sin ( x )}

= \frac{\frac{2 s i n ( x ) + 1}{s i n ( x )} cos ( x )}{1}

Apply the fraction rule:

\frac{a}{1} = a

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

Multiply fractions:

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

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

Expand

(2 sin (x) + 1) cos (x) - 2 cos (x) sin (x) : cos (x)

(2 sin (x) + 1) cos (x) - 2 cos (x) sin (x)

= cos (x) (2 sin (x) + 1) - 2 cos (x) sin (x)

Expand

cos (x) (2 sin (x) + 1) : 2 cos (x) sin (x) + cos (x)

cos (x) (2 sin (x) + 1)

Apply the distributive law:

a (b + c) = ab + a c

a = cos (x), b = 2 sin (x), c = 1

= cos (x) \cdot 2 sin (x) + cos (x) \cdot 1

= 2 cos (x) sin (x) + 1 \cdot cos (x)

Multiply:

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

= 2 cos (x) sin (x) + cos (x)

= 2 cos (x) sin (x) + cos (x) - 2 cos (x) sin (x)

Add similar elements:

2 cos (x) sin (x) - 2 cos (x) sin (x) = 0

= cos (x)

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

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

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

Use the basic trigonometric identity:

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

= cot (x)

We showed that the two sides could take the same form

\Rightarrow True

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

1 - \frac{( cos ^{2} ( x ) )}{1 + sin ( x )} = sin (x)

- sin^{2} (x) + cos^{2} (x) = cos (2 x)

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

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

Is (2+csc(x))/(sec(x))-2cos(x)=cot(x) ?
The answer to whether (2+csc(x))/(sec(x))-2cos(x)=cot(x) is True