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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply the fraction rule:

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

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

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

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

Apply exponent rule:

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

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

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

Multiply

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos (x)

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

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

Apply the fraction rule:

\frac{a}{1} = a

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

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

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

Rewrite using trig identities

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

Use the basic trigonometric identity:

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

sin (x) tan (x)

sin (x) tan (x)

We showed that the two sides could take the same form

\Rightarrow True

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

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

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

p ro v e cot (x) (sec^{2} (x) - 1) = tan (x)

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

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