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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Use the negative angle identity:

sin (- x) = - sin (x)

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

(- sin (x))^{2}

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

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

= sin^{2} (x)

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

Apply the fraction rule:

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

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

Cancel the common factor:

sin^{2} (x)

= cos^{2} (x)

= cos^{2} (x)

= cos^{2} (x)

We showed that the two sides could take the same form

\Rightarrow True

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

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

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

sin (x) cos (x) + sin^{3} (x) sec (x) = tan (x)

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

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