Is ((cos(x)sec(x)))/(tan(x))=cot(x) ?

The answer to whether ((cos(x)sec(x)))/(tan(x))=cot(x) is True

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prove ((cos(x)sec(x)))/(tan(x))=cot(x)

prove

\frac{( cos ( x ) sec ( x ) )}{tan ( x )} = cot (x)

True

Solution steps

\frac{cos ( x ) sec ( x )}{tan ( x )} = cot (x)

Manipulating left side

\frac{cos ( x ) sec ( x )}{tan ( x )}

Express with sin, cos

\frac{cos ( x ) sec ( x )}{tan ( x )}

Use the basic trigonometric identity:

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

= \frac{cos ( x ) \frac{1}{c o s ( x )}}{tan ( x )}

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply the fraction rule:

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

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

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

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

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

cos (x) cos (x) = cos^{1 + 1} (x)

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

Add the numbers:

1 + 1 = 2

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

= \frac{\frac{1}{c o s ( x )} cos ^{2} ( x )}{sin ( x )}

Multiply

\frac{1}{cos ( x )} cos^{2} (x) : cos (x)

\frac{1}{cos ( x )} cos^{2} (x)

Multiply fractions:

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

= \frac{1 \cdot cos ^{2} ( x )}{cos ( x )}

Multiply:

1 \cdot cos^{2} (x) = cos^{2} (x)

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

Cancel the common factor:

cos (x)

= 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

p ro v e (sin^{2} (x) + cos^{2} (x))^{5} = 1

p ro v e \frac{cot ^{2} ( x ) - 1}{2 cot ( x )} = cot (2 x)

p ro v e csc (θ) = \frac{sec ( θ )}{tan ( θ )}

p ro v e \frac{cos ( 2 x ) + 1}{sin ( 2 x )} = cot (x)

p ro v e - tan (t) - \frac{cos ( t )}{sin ( t ) - 1} = sec (x)

Is ((cos(x)sec(x)))/(tan(x))=cot(x) ?
The answer to whether ((cos(x)sec(x)))/(tan(x))=cot(x) is True