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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply exponent rule:

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

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

Join

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

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

Convert element to fraction:

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

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

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

Multiply:

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

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos^{2} (y)

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

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

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

Rewrite using trig identities

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

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

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

sec (A) - cos (A) = tan (A) sin (A)

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

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

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