Is (1+tan(θ))^2=sec^2(θ)+2tan(θ) ?

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

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prove (1+tan(θ))^2=sec^2(θ)+2tan(θ)

prove

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

True

Solution steps

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

Manipulating left side

(1 + tan (θ))^{2}

Expand

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

(1 + tan (θ))^{2}

Apply Perfect Square Formula:

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

a = 1, b = tan (θ)

= 1^{2} + 2 \cdot 1 \cdot tan (θ) + tan^{2} (θ)

Simplify

1^{2} + 2 \cdot 1 \cdot tan (θ) + tan^{2} (θ) : 1 + 2 tan (θ) + tan^{2} (θ)

1^{2} + 2 \cdot 1 \cdot tan (θ) + tan^{2} (θ)

Apply rule

1^{a} = 1

1^{2} = 1

= 1 + 2 \cdot 1 \cdot tan (θ) + tan^{2} (θ)

Multiply the numbers:

2 \cdot 1 = 2

= 1 + 2 tan (θ) + tan^{2} (θ)

= 1 + 2 tan (θ) + tan^{2} (θ)

= 1 + tan^{2} (θ) + 2 tan (θ)

Manipulating right side

sec^{2} (θ) + 2 tan (θ)

Rewrite using trig identities

sec^{2} (θ) + 2 tan (θ)

Use the Pythagorean identity:

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

= tan^{2} (θ) + 1 + 2 tan (θ)

= tan^{2} (θ) + 1 + 2 tan (θ)

We showed that the two sides could take the same form

\Rightarrow True

\frac{1 + cot ^{2} ( x )}{2 cot ( x )} = csc (2 x)

sin^{2} (x) + tan^{2} (x) sin^{2} (x) = tan^{2} (x)

\frac{sin ( θ )}{1 - sin ^{2} ( θ )} = sec (θ) tan (θ)

sin^{2} (θ) cot^{2} (θ) = 1

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

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