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

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

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prove (sec(e)-1)(sec(e)+1)=tan^2(e)

prove

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

True

Solution steps

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

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

= 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)

\Rightarrow True

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

\frac{( tan ( x ) cos ( x ) )}{sin ( x )} = 1

tan (\frac{π}{3}) sin (\frac{2 π}{3}) = 1 + sin (\frac{π}{6})

2 cot (x) + cos (x) - sec (x) = cot (x)

(tan (θ) + cot (θ)) sin (θ) cos (θ) = 1

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