Is (1-sin^2(x))/(cot^2(x))=1-cos^2(x) ?

The answer to whether (1-sin^2(x))/(cot^2(x))=1-cos^2(x) is True

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prove (1-sin^2(x))/(cot^2(x))=1-cos^2(x)

prove

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

True

Solution steps

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

Manipulating left side

\frac{1 - sin ^{2} ( x )}{cot ^{2} ( x )}

Express with sin, cos

\frac{1 - sin ^{2} ( x )}{cot ^{2} ( x )}

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

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

Apply exponent rule:

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

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

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

Apply the fraction rule:

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

cos^{2} (x) + sin^{2} (x) = 1

sin^{2} (x) = 1 - cos^{2} (x)

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

Simplify

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

\frac{( 1 - ( 1 - cos ^{2} ( x ) ) ) ( 1 - cos ^{2} ( x ) )}{cos ^{2} ( x )}

Expand

1 - (1 - cos^{2} (x)) : cos^{2} (x)

1 - (1 - cos^{2} (x))

- (1 - cos^{2} (x)) : - 1 + cos^{2} (x)

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

Distribute parentheses

= - (1) - (- cos^{2} (x))

Apply minus-plus rules

- (- a) = a, - (a) = - a

= - 1 + cos^{2} (x)

= 1 - 1 + cos^{2} (x)

1 - 1 = 0

= cos^{2} (x)

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

Cancel the common factor:

cos^{2} (x)

= 1 - cos^{2} (x)

= 1 - cos^{2} (x)

= 1 - cos^{2} (x)

We showed that the two sides could take the same form

\Rightarrow True

(1 + sin (x)) (sec (x) - tan (x)) = cos (x)

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

\frac{sin ( θ ) sec ( θ )}{tan ( θ )} = 1

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

tan (\frac{π}{6}) = \frac{1}{3}

Is (1-sin^2(x))/(cot^2(x))=1-cos^2(x) ?
The answer to whether (1-sin^2(x))/(cot^2(x))=1-cos^2(x) is True