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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Multiply fractions:

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

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

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

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

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

Cancel the common factor:

cos^{2} (x)

= sin^{2} (x)

= sin^{2} (x)

We showed that the two sides could take the same form

\Rightarrow True

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

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

cos (θ) sec (θ) = 1

cos (a + b) + cos (a - b) = 2 cos (a) cos (b)

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

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