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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

Simplify

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

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

Expand

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

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

Apply the distributive law:

a (b + c) = ab + a c

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

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

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

Simplify

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

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

tan^{2} (x) tan^{2} (x) = tan^{4} (x)

tan^{2} (x) tan^{2} (x)

Apply exponent rule:

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

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

= tan^{2 + 2} (x)

Add the numbers:

2 + 2 = 4

= tan^{4} (x)

1 \cdot tan^{2} (x) = tan^{2} (x)

1 \cdot tan^{2} (x)

Multiply:

1 \cdot tan^{2} (x) = tan^{2} (x)

= tan^{2} (x)

= tan^{4} (x) + tan^{2} (x)

= tan^{4} (x) + tan^{2} (x)

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

Add similar elements:

tan^{2} (x) - tan^{2} (x) = 0

= tan^{4} (x)

= tan^{4} (x)

= tan^{4} (x)

We showed that the two sides could take the same form

\Rightarrow True

sin^{4} (x) + cos^{4} (x) = \frac{3}{4} + \frac{1}{4} cos (4 x)

sin^{2} (x) + 3 cos^{2} (x) = 1 + 2 cos^{2} (x)

sec (2 θ) = \frac{csc ^{2} ( θ )}{csc ^{2} ( θ ) - 2}

sin (x + \frac{3 π}{2}) = - cos (x)

sec (x) - sec (x) \cdot sin^{2} (x) = cos (x)

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