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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply exponent rule:

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

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

Join

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

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

Convert element to fraction:

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

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

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

Multiply:

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

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos^{2} (A)

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

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

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

Rewrite using trig identities

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

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

sin^{4} (θ) + cos^{2} (θ) sin^{2} (θ) = sin^{2} (θ)

cos (10 x) - cos (6 x) = - 2 sin (8 x) sin (2 x)

cot (2 x) = \frac{tan ( x ) ( cot ^{2} ( x ) - 1 )}{2}

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

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