Is (1+tan(a))*(1-tan(a))=2-sec^2(a) ?

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

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prove (1+tan(a))*(1-tan(a))=2-sec^2(a)

prove

(1 + tan (a)) \cdot (1 - tan (a)) = 2 - sec^{2} (a)

True

Solution steps

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

Manipulating left side

(1 + tan (a)) (1 - tan (a))

Expand

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

(1 + tan (a)) (1 - tan (a))

Apply Difference of Two Squares Formula:

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

a = 1, b = tan (a)

= 1^{2} - tan^{2} (a)

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - tan^{2} (a)

= 1 - tan^{2} (a)

Rewrite using trig identities

1 - tan^{2} (a)

Use the Pythagorean identity:

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

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

= 1 - (sec^{2} (a) - 1)

Simplify

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

1 - (sec^{2} (a) - 1)

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

- (sec^{2} (a) - 1)

Distribute parentheses

= - (sec^{2} (a)) - (- 1)

Apply minus-plus rules

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

= - sec^{2} (a) + 1

= 1 - sec^{2} (a) + 1

Simplify

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

1 - sec^{2} (a) + 1

Group like terms

= - sec^{2} (a) + 1 + 1

Add the numbers:

1 + 1 = 2

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

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

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

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

= 2 - sec^{2} (a)

We showed that the two sides could take the same form

\Rightarrow True

(cot (x)) (sec (x)) (sin (x)) = 1

\frac{cos ^{2} ( x ) - 4}{cos ( x ) - 2} = cos (x) + 2

sin^{2} (a) tan^{2} (a) = tan^{2} (a) - sin^{2} (a)

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

tan (θ) (sin (θ) + cot (θ) cos (θ)) = sec (θ)

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