Is (cot^2(θ))/(cos^2(θ))=csc^2(θ) ?

The answer to whether (cot^2(θ))/(cos^2(θ))=csc^2(θ) is True

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prove (cot^2(θ))/(cos^2(θ))=csc^2(θ)

prove

\frac{cot ^{2} ( θ )}{cos ^{2} ( θ )} = csc^{2} (θ)

True

Solution steps

\frac{cot ^{2} ( θ )}{cos ^{2} ( θ )} = csc^{2} (θ)

Manipulating left side

\frac{cot ^{2} ( θ )}{cos ^{2} ( θ )}

Express with sin, cos

\frac{cot ^{2} ( θ )}{cos ^{2} ( θ )}

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

(\frac{cos ( θ )}{sin ( θ )})^{2}

Apply exponent rule:

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

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

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

Apply the fraction rule:

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

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

Cancel the common factor:

cos^{2} (θ)

= \frac{1}{sin ^{2} ( θ )}

= \frac{1}{sin ^{2} ( θ )}

= \frac{1}{sin ^{2} ( θ )}

Rewrite using trig identities

Use the basic trigonometric identity:

sin (x) = \frac{1}{csc ( x )}

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

Simplify

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

(\frac{1}{csc ( θ )})^{2} = \frac{1}{csc ^{2} ( θ )}

(\frac{1}{csc ( θ )})^{2}

Apply exponent rule:

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

= \frac{1 ^{2}}{csc ^{2} ( θ )}

Apply rule

1^{a} = 1

1^{2} = 1

= \frac{1}{csc ^{2} ( θ )}

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

Apply the fraction rule:

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

= \frac{csc ^{2} ( θ )}{1}

Apply rule

\frac{a}{1} = a

= csc^{2} (θ)

csc^{2} (θ)

csc^{2} (θ)

We showed that the two sides could take the same form

\Rightarrow True

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

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

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

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

25 cos^{2} (x) = 25 - 25 sin^{2} (x)

Is (cot^2(θ))/(cos^2(θ))=csc^2(θ) ?
The answer to whether (cot^2(θ))/(cos^2(θ))=csc^2(θ) is True