Is (1-sin^2(θ))sec^2(θ)=1 ?

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

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prove (1-sin^2(θ))sec^2(θ)=1

prove

(1 - sin^{2} (θ)) sec^{2} (θ) = 1

True

Solution steps

(1 - sin^{2} (θ)) sec^{2} (θ) = 1

Manipulating left side

(1 - sin^{2} (θ)) sec^{2} (θ)

Express with sin, cos

(1 - sin^{2} (θ)) sec^{2} (θ)

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

(\frac{1}{cos ( θ )})^{2}

Apply exponent rule:

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

= \frac{1 ^{2}}{cos ^{2} ( θ )}

Apply rule

1^{a} = 1

1^{2} = 1

= \frac{1}{cos ^{2} ( θ )}

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

Multiply fractions:

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

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

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

1 \cdot (1 - sin^{2} (θ))

Multiply:

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

= (1 - sin^{2} (θ))

Remove parentheses:

(a) = a

= 1 - sin^{2} (θ)

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Apply rule

\frac{a}{a} = 1

= 1

= 1

We showed that the two sides could take the same form

\Rightarrow True

(1 + sin (a)) (1 - sin (a)) = cos^{2} (a)

sin^{3} (θ) = \frac{3}{4} sin (θ) - \frac{1}{4} sin (3 θ)

tan^{2} (θ) + 1 = sec^{2} (θ)

sin (x) + sin (3 x) = 4 sin (x) cos^{2} (x)

\frac{cos ( x ) sec ( x )}{tan ( x )} = cot (x)

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