Is tan^4(θ)=tan^2(θ)sec^2(θ)-tan^2(θ) ?

The answer to whether tan^4(θ)=tan^2(θ)sec^2(θ)-tan^2(θ) is True

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prove tan^4(θ)=tan^2(θ)sec^2(θ)-tan^2(θ)

prove

tan^{4} (θ) = tan^{2} (θ) sec^{2} (θ) - tan^{2} (θ)

True

Solution steps

tan^{4} (θ) = tan^{2} (θ) sec^{2} (θ) - tan^{2} (θ)

Manipulating right side

tan^{2} (θ) sec^{2} (θ) - tan^{2} (θ)

Rewrite using trig identities

tan^{2} (θ) sec^{2} (θ) - tan^{2} (θ)

Use the Pythagorean identity:

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

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

Simplify

tan^{2} (θ) (tan^{2} (θ) + 1) - tan^{2} (θ) : tan^{4} (θ)

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

Expand

tan^{2} (θ) (tan^{2} (θ) + 1) : tan^{4} (θ) + tan^{2} (θ)

tan^{2} (θ) (tan^{2} (θ) + 1)

Apply the distributive law:

a (b + c) = ab + a c

a = tan^{2} (θ), b = tan^{2} (θ), c = 1

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

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

Simplify

tan^{2} (θ) tan^{2} (θ) + 1 \cdot tan^{2} (θ) : tan^{4} (θ) + tan^{2} (θ)

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

tan^{2} (θ) tan^{2} (θ) = tan^{4} (θ)

tan^{2} (θ) tan^{2} (θ)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

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

= tan^{2 + 2} (θ)

Add the numbers:

2 + 2 = 4

= tan^{4} (θ)

1 \cdot tan^{2} (θ) = tan^{2} (θ)

1 \cdot tan^{2} (θ)

Multiply:

1 \cdot tan^{2} (θ) = tan^{2} (θ)

= tan^{2} (θ)

= tan^{4} (θ) + tan^{2} (θ)

= tan^{4} (θ) + tan^{2} (θ)

= tan^{4} (θ) + tan^{2} (θ) - tan^{2} (θ)

Add similar elements:

tan^{2} (θ) - tan^{2} (θ) = 0

= tan^{4} (θ)

= tan^{4} (θ)

= tan^{4} (θ)

We showed that the two sides could take the same form

\Rightarrow True

\frac{sin ( 6 A ) + sin ( 2 A )}{cos ( 6 A ) - cos ( 2 A )} = - cot (2 A)

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

cot (x) - sin (2 x) = cot (x) cos (2 x)

sin (2 θ) = 2 sin (2 θ) cos (2 θ)

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

Is tan^4(θ)=tan^2(θ)sec^2(θ)-tan^2(θ) ?
The answer to whether tan^4(θ)=tan^2(θ)sec^2(θ)-tan^2(θ) is True