Is sec(x)*(tan(x))/(sin(x))=sec^2(x) ?

The answer to whether sec(x)*(tan(x))/(sin(x))=sec^2(x) is True

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prove sec(x)*(tan(x))/(sin(x))=sec^2(x)

prove

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

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

= \frac{\frac{s i n ( x )}{c o s ( x )}}{sin ( x )} sec (x)

Use the basic trigonometric identity:

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

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

Simplify

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

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

Multiply fractions:

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

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

Multiply:

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

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

Apply the fraction rule:

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

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

Cancel the common factor:

sin (x)

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

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

cos (x) cos (x)

Apply exponent rule:

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

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

= cos^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= cos^{2} (x)

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

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

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

Rewrite using trig identities

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

Apply exponent rule:

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

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

Apply the fraction rule:

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

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

Apply rule

\frac{a}{1} = a

= sec^{2} (x)

sec^{2} (x)

sec^{2} (x)

We showed that the two sides could take the same form

\Rightarrow True

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

\frac{1}{sin ( θ )} - \frac{1}{tan ( θ )} = tan (\frac{θ}{2})

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

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

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

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