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

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

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prove 1/(sin^2(x))-1/(tan^2(x))=1

prove

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

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

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

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

Apply the fraction rule:

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

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

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

Apply rule

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Apply rule

\frac{a}{a} = 1

= 1

= 1

We showed that the two sides could take the same form

\Rightarrow True

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

cos (x + π) + sin (x - \frac{3 π}{2}) = 0

cos^{4} (A) - sin^{4} (A) + 1 = 2 cos^{2} (A)

- cot (θ) cos (θ) - sin (θ) = - cot (θ) sec (θ)

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

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