Is cot(x)sin(x)sec^2(x)=sec(x) ?

The answer to whether cot(x)sin(x)sec^2(x)=sec(x) is True

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prove cot(x)sin(x)sec^2(x)=sec(x)

prove

cot (x) sin (x) sec^{2} (x) = sec (x)

True

Solution steps

cot (x) sin (x) sec^{2} (x) = sec (x)

Manipulating left side

cot (x) sin (x) sec^{2} (x)

Express with sin, cos

cot (x) sec^{2} (x) sin (x)

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Multiply fractions:

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

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

Cancel the common factor:

sin (x)

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

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

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

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

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

Apply rule

1^{a} = 1

1^{2} = 1

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

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

Multiply fractions:

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

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

Multiply:

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

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

Cancel the common factor:

cos (x)

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

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

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

Rewrite using trig identities

Use the basic trigonometric identity:

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

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

Simplify

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

Apply the fraction rule:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

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

Apply rule

\frac{a}{1} = a

= sec (x)

sec (x)

sec (x)

We showed that the two sides could take the same form

\Rightarrow True

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

sin^{2} (x) + 4 cos^{2} (x) = 4 - 3 sin^{2} (x)

sin (a + b) = sin (a) cos (b) + cos (a) sin (b)

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

tan (θ + \frac{π}{2}) = - cot (θ)

Is cot(x)sin(x)sec^2(x)=sec(x) ?
The answer to whether cot(x)sin(x)sec^2(x)=sec(x) is True