Is tan^2(x)-(sec(x)+1)(sec(x)-1)=0 ?

The answer to whether tan^2(x)-(sec(x)+1)(sec(x)-1)=0 is True

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prove tan^2(x)-(sec(x)+1)(sec(x)-1)=0

prove

tan^{2} (x) - (sec (x) + 1) (sec (x) - 1) = 0

True

Solution steps

tan^{2} (x) - (sec (x) + 1) (sec (x) - 1) = 0

Manipulating left side

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

Rewrite using trig identities

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

Expand

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

(sec (x) + 1) (sec (x) - 1)

Apply Difference of Two Squares Formula:

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

a = sec (x), b = 1

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

Apply rule

1^{a} = 1

1^{2} = 1

= sec^{2} (x) - 1

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

Use the Pythagorean identity:

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

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

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

Add similar elements:

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

= 0

= 0

We showed that the two sides could take the same form

\Rightarrow True

p ro v e sec^{4} (θ) = sec^{2} (θ) tan^{2} (θ) + sec^{2} (θ)

p ro v e sin^{2} (θ) = 1 - \frac{1}{sec ^{2} ( θ )}

p ro v e tan (θ) sec (θ) csc (θ) = 1 + tan^{2} (θ)

p ro v e \frac{1 - cos ( x )}{\frac{1}{c o s ( x )} - 1} = cos (x)

p ro v e sin (π - 0) = sin (0)

Is tan^2(x)-(sec(x)+1)(sec(x)-1)=0 ?
The answer to whether tan^2(x)-(sec(x)+1)(sec(x)-1)=0 is True