Is (1-sin^2(x))sec^2(x)=1 ?

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

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prove (1-sin^2(x))sec^2(x)=1

prove

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

True

Solution steps

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

Manipulating left side

(1 - sin^{2} (x)) sec^{2} (x)

Express with sin, cos

(1 - sin^{2} (x)) sec^{2} (x)

Use the basic trigonometric identity:

sec (x) = \frac{1}{cos ( x )}

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

Simplify

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

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

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

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

Apply exponent rule:

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

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

Multiply fractions:

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

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

1 \cdot (1 - sin^{2} (x)) = 1 - sin^{2} (x)

1 \cdot (1 - sin^{2} (x))

Multiply:

1 \cdot (1 - sin^{2} (x)) = (1 - sin^{2} (x))

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

Remove parentheses:

(a) = a

= 1 - sin^{2} (x)

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Apply rule

\frac{a}{a} = 1

= 1

= 1

We showed that the two sides could take the same form

\Rightarrow True

tan (x) sin (x) = 1 - cos (x)

\frac{tan ( x )}{sec ( x ) - cos ( x )} = csc (x)

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

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

tan (2 θ) = \frac{2 tan ( θ )}{( 1 - tan ( 2 θ ) )}

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