Is (1+tan^2(α))(1-sin^2(α))=1 ?

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

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prove (1+tan^2(α))(1-sin^2(α))=1

prove

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

True

Solution steps

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

Manipulating left side

(1 + tan^{2} (α)) (1 - sin^{2} (α))

Rewrite using trig identities

(1 + tan^{2} (α)) (1 - sin^{2} (α))

Use the Pythagorean identity:

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

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

= (1 + tan^{2} (α)) cos^{2} (α)

= (1 + tan^{2} (α)) cos^{2} (α)

Express with sin, cos

(1 + tan^{2} (α)) cos^{2} (α)

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply exponent rule:

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

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

Join

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

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

Convert element to fraction:

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

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

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

Multiply:

1 \cdot cos^{2} (α) = cos^{2} (α)

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos^{2} (α)

= cos^{2} (α) + sin^{2} (α)

= cos^{2} (α) + sin^{2} (α)

= cos^{2} (α) + sin^{2} (α)

Rewrite using trig identities

cos^{2} (α) + sin^{2} (α)

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

p ro v e csc (θ) - cot (θ) = \frac{1 - cos ( θ )}{sin ( θ )}

p ro v e tan (x) = tan (x) + 2 sin^{2} (x)

p ro v e \frac{1}{2 cot ( 1 - cos ^{2} ( x ) )} = csc (2 x)

p ro v e cot (u) = \frac{1}{tan ( u )}

p ro v e - sin (2 x) = - 4 cos (x) sin (x)

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