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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

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

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

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

Expand

sec^{2} (x) (sec^{2} (x) - 1) : sec^{4} (x) - sec^{2} (x)

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

Apply the distributive law:

a (b - c) = ab - a c

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

= sec^{2} (x) sec^{2} (x) - sec^{2} (x) \cdot 1

= sec^{2} (x) sec^{2} (x) - 1 \cdot sec^{2} (x)

Simplify

sec^{2} (x) sec^{2} (x) - 1 \cdot sec^{2} (x) : sec^{4} (x) - sec^{2} (x)

sec^{2} (x) sec^{2} (x) - 1 \cdot sec^{2} (x)

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

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

Apply exponent rule:

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

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

= sec^{2 + 2} (x)

Add the numbers:

2 + 2 = 4

= sec^{4} (x)

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

1 \cdot sec^{2} (x)

Multiply:

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

= sec^{2} (x)

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

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

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

Add similar elements:

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

= sec^{4} (x)

= sec^{4} (x)

= sec^{4} (x)

We showed that the two sides could take the same form

\Rightarrow True

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