Is (cos(θ))^2(sec(θ))^2=1 ?

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

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prove (cos(θ))^2(sec(θ))^2=1

prove

(cos (θ))^{2} (sec (θ))^{2} = 1

True

Solution steps

(cos (θ))^{2} (sec (θ))^{2} = 1

Manipulating left side

(cos (θ))^{2} (sec (θ))^{2}

Simplify

= cos^{2} (θ) sec^{2} (θ)

Express with sin, cos

cos^{2} (θ) sec^{2} (θ)

Use the basic trigonometric identity:

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

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

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

cos^{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 )} cos^{2} (θ)

Multiply fractions:

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

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

Cancel the common factor:

cos^{2} (θ)

= 1

= 1

= 1

We showed that the two sides could take the same form

\Rightarrow True

p ro v e sin (β) = \frac{1}{4}

p ro v e 3 - 6 sin^{2} (x) = 0

p ro v e cos^{4} (θ) = \frac{1}{4} + \frac{cos ( 2 θ )}{2} + \frac{cos ^{2} ( 2 θ )}{4}

p ro v e sin (x) (tan^{2} (x) + 1) = tan (x) sec (x)

p ro v e csc (u) = 2

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