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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply exponent rule:

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

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

Join

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

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

Convert element to fraction:

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

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

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

Multiply:

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

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos^{2} (t)

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

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

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

Rewrite using trig identities

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

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

1 - cosh^{2} (x) = - sinh^{2} (x)

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

cot^{2} (θ) + 1 = csc^{2} (θ)

2 cos (θ) tan (θ) csc (θ) = 2

sec (a) - sec (a) sin^{2} (a) = cos (a)

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