Is (tan^2(θ))/(sec(θ))=sin(θ)tan(θ) ?

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

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

prove

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

True

Solution steps

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

Manipulating left side

\frac{tan ^{2} ( θ )}{sec ( θ )}

Express with sin, cos

\frac{tan ^{2} ( θ )}{sec ( θ )}

Use the basic trigonometric identity:

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

= \frac{( \frac{s i n ( θ )}{c o s ( θ )} ) ^{2}}{sec ( θ )}

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply the fraction rule:

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

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

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

(\frac{sin ( θ )}{cos ( θ )})^{2}

Apply exponent rule:

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

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

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

Multiply

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos (θ)

= \frac{sin ^{2} ( θ )}{cos ( θ )}

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

Apply the fraction rule:

\frac{a}{1} = a

= \frac{sin ^{2} ( θ )}{cos ( θ )}

= \frac{sin ^{2} ( θ )}{cos ( θ )}

= \frac{sin ^{2} ( θ )}{cos ( θ )}

Rewrite using trig identities

= sin (θ) \frac{sin ( θ )}{cos ( θ )}

Use the basic trigonometric identity:

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

sin (θ) tan (θ)

sin (θ) tan (θ)

We showed that the two sides could take the same form

\Rightarrow True

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

\frac{tan ^{2} ( θ )}{sec ( θ )} = sec (θ) - cos (θ)

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

2 csc (2 x) = tan (x) + cot (x)

cos (x + y) = cos (x) cos (y) - sin (x) sin (y)

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