Is (tan^2(θ))/(sin(θ))=sec(θ)tan(θ) ?

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

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

prove

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

True

Solution steps

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

Manipulating left side

\frac{tan ^{2} ( θ )}{sin ( θ )}

Express with sin, cos

\frac{tan ^{2} ( θ )}{sin ( θ )}

Use the basic trigonometric identity:

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

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

Simplify

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

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

(\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 )}}{sin ( θ )}

Apply the fraction rule:

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

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

Cancel the common factor:

sin (θ)

= \frac{sin ( θ )}{cos ^{2} ( θ )}

= \frac{sin ( θ )}{cos ^{2} ( θ )}

= \frac{sin ( θ )}{cos ^{2} ( θ )}

Manipulating right side

sec (θ) tan (θ)

Express with sin, cos

sec (θ) tan (θ)

Use the basic trigonometric identity:

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

= \frac{1}{cos ( θ )} tan (θ)

Use the basic trigonometric identity:

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

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

Simplify

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

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

Multiply fractions:

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

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

Multiply:

1 \cdot sin (θ) = sin (θ)

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

cos (θ) cos (θ) = cos^{2} (θ)

cos (θ) cos (θ)

Apply exponent rule:

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

cos (θ) cos (θ) = cos^{1 + 1} (θ)

= cos^{1 + 1} (θ)

Add the numbers:

1 + 1 = 2

= cos^{2} (θ)

= \frac{sin ( θ )}{cos ^{2} ( θ )}

= \frac{sin ( θ )}{cos ^{2} ( θ )}

= \frac{sin ( θ )}{cos ^{2} ( θ )}

We showed that the two sides could take the same form

\Rightarrow True

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

sin (2 x) + sin (2 y) = 2 sin (x + y) cos (x - y)

sin (θ) csc (θ) cos (θ) = cos (θ)

cot (x) - csc^{2} (x) cot (x) = - cot^{3} (x)

cot (x - \frac{π}{2}) = - tan (x)

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