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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

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

Express with sin, cos

(1 + tan^{2} (x)) cos^{2} (x)

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply exponent rule:

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

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

Join

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

Multiply:

1 \cdot cos^{2} (x) = cos^{2} (x)

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos^{2} (x)

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

= 1

= 1

We showed that the two sides could take the same form

\Rightarrow True

\frac{cos ( θ )}{1 - sin ( θ )} = \frac{sin ( θ ) - csc ( θ )}{cos ( θ ) - cot ( θ )}

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

(sec^{2} (θ) - 1) cos^{2} (θ) = sin^{2} (θ)

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

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

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