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

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

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

prove

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

True

Solution steps

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

Manipulating right side

sec^{2} (θ) tan^{2} (θ) + sec^{2} (θ)

Rewrite using trig identities

sec^{2} (θ) tan^{2} (θ) + sec^{2} (θ)

Use the Pythagorean identity:

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

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

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

Simplify

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

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

Expand

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

sec^{2} (θ) (sec^{2} (θ) - 1)

Apply the distributive law:

a (b - c) = ab - a c

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

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

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

Simplify

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

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

sec^{2} (θ) sec^{2} (θ) = sec^{4} (θ)

sec^{2} (θ) sec^{2} (θ)

Apply exponent rule:

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

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

= sec^{2 + 2} (θ)

Add the numbers:

2 + 2 = 4

= sec^{4} (θ)

1 \cdot sec^{2} (θ) = sec^{2} (θ)

1 \cdot sec^{2} (θ)

Multiply:

1 \cdot sec^{2} (θ) = sec^{2} (θ)

= sec^{2} (θ)

= sec^{4} (θ) - sec^{2} (θ)

= sec^{4} (θ) - sec^{2} (θ)

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

Add similar elements:

- sec^{2} (θ) + sec^{2} (θ) = 0

= sec^{4} (θ)

= sec^{4} (θ)

= sec^{4} (θ)

We showed that the two sides could take the same form

\Rightarrow True

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

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

\frac{1 - cos ( x )}{\frac{1}{c o s ( x )} - 1} = cos (x)

sin (π - 0) = sin (0)

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