Is (tan(θ)+1)(tan(θ)-1)=sec^2(θ)-2 ?

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

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prove (tan(θ)+1)(tan(θ)-1)=sec^2(θ)-2

prove

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

True

Solution steps

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

Manipulating left side

(tan (θ) + 1) (tan (θ) - 1)

Expand

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

(- 1 + tan (θ)) (1 + tan (θ))

Apply Difference of Two Squares Formula:

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

a = tan (θ), b = 1

= tan^{2} (θ) - 1^{2}

Apply rule

1^{a} = 1

1^{2} = 1

= tan^{2} (θ) - 1

= - 1 + tan^{2} (θ)

Manipulating right side

sec^{2} (θ) - 2

Rewrite using trig identities

sec^{2} (θ) - 2

Use the Pythagorean identity:

sec^{2} (x) = tan^{2} (x) + 1

= tan^{2} (θ) + 1 - 2

Simplify

= tan^{2} (θ) - 1

= tan^{2} (θ) - 1

We showed that the two sides could take the same form

\Rightarrow True

\frac{sec ( t ) - 1}{1 - cos ( t )} = sec (t)

tan (\frac{7 π}{6}) = \frac{3}{3}

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

- tan (θ) - \frac{cos ( θ )}{sin ( θ ) - 1} = sec (θ)

cot (θ) \cdot sec (θ) = csc (θ)

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