Is (1+tan(β))(1-tan(β))=2-sec^2(β) ?

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

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prove (1+tan(β))(1-tan(β))=2-sec^2(β)

prove

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

True

Solution steps

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

Manipulating left side

(1 + tan (β)) (1 - tan (β))

Expand

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

(1 + tan (β)) (1 - tan (β))

Apply Difference of Two Squares Formula:

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

a = 1, b = tan (β)

= 1^{2} - tan^{2} (β)

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - tan^{2} (β)

= 1 - tan^{2} (β)

Rewrite using trig identities

1 - tan^{2} (β)

Use the Pythagorean identity:

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

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

= 1 - (sec^{2} (β) - 1)

Simplify

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

1 - (sec^{2} (β) - 1)

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

- (sec^{2} (β) - 1)

Distribute parentheses

= - (sec^{2} (β)) - (- 1)

Apply minus-plus rules

- (- a) = a, - (a) = - a

= - sec^{2} (β) + 1

= 1 - sec^{2} (β) + 1

Simplify

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

1 - sec^{2} (β) + 1

Group like terms

= - sec^{2} (β) + 1 + 1

Add the numbers:

1 + 1 = 2

= - sec^{2} (β) + 2

= - sec^{2} (β) + 2

= - sec^{2} (β) + 2

= - sec^{2} (β) + 2

= 2 - sec^{2} (β)

We showed that the two sides could take the same form

\Rightarrow True

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

cos^{2} (x) - cos^{2} (9 0^{\circ} - x) = 1 - 2 sin (x)

sec^{2} (t) csc^{2} (t) = sec^{2} (t) + csc^{2} (t)

\frac{1}{1 + sin ( x )} + \frac{1}{1 + csc ( x )} = 1

sin^{2} (θ) cos^{2} (θ) = 1

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