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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Multiply

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

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

Multiply fractions:

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

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

Multiply:

1 \cdot sin (x) = sin (x)

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

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

Divide fractions:

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

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

Refine

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

Cancel the common factor:

cos^{2} (x)

= sin (x)

= sin (x)

= sin (x)

We showed that the two sides could take the same form

\Rightarrow True

sin (α) = \frac{1}{csc ( α )}

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

5 sec^{2} (x) + 4 = 9 sec^{2} (x) - 4 tan^{2} (x)

csc (- x) - sin (- x) = cos (x) cot (x)

cos (20 π - x) + sin (\frac{83 π}{2} + x) = 0

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