Is (1+sec(θ))(1-sec(θ))=-tan^2(θ) ?

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

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prove (1+sec(θ))(1-sec(θ))=-tan^2(θ)

prove

(1 + sec (θ)) (1 - sec (θ)) = - tan^{2} (θ)

True

Solution steps

(1 + sec (θ)) (1 - sec (θ)) = - tan^{2} (θ)

Manipulating left side

(1 + sec (θ)) (1 - sec (θ))

Expand

(1 + sec (θ)) (1 - sec (θ)) : 1 - sec^{2} (θ)

(1 + sec (θ)) (1 - sec (θ))

Apply Difference of Two Squares Formula:

(a + b) (a - b) = a^{2} - b^{2}

a = 1, b = sec (θ)

= 1^{2} - sec^{2} (θ)

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - sec^{2} (θ)

= 1 - sec^{2} (θ)

Use the Pythagorean identity:

sec^{2} (θ) = tan^{2} (θ) + 1

sec^{2} (θ) - 1 = tan^{2} (θ)

= - tan^{2} (θ)

We showed that the two sides could take the same form

\Rightarrow True

cos (x) = \frac{cot ( x )}{csc ( x )}

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

cot (t) cos (t) = csc (t) - sin (t)

sin (\frac{3 π}{2} - x) = - cos (x)

sin (x) sec (x) = \frac{1}{cot ( x )}

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