Is (1-4sec^2(x))/(1+2sec(x))=1-2sec(x) ?

The answer to whether (1-4sec^2(x))/(1+2sec(x))=1-2sec(x) is True

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prove (1-4sec^2(x))/(1+2sec(x))=1-2sec(x)

prove

\frac{1 - 4 sec ^{2} ( x )}{1 + 2 sec ( x )} = 1 - 2 sec (x)

True

Solution steps

\frac{1 - 4 sec ^{2} ( x )}{1 + 2 sec ( x )} = 1 - 2 sec (x)

Manipulating left side

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

Simplify

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

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

Factor

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

1 - 4 sec^{2} (x)

Factor out common term

- 1

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

Factor

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

4 sec^{2} (x) - 1

Rewrite

4 sec^{2} (x) - 1

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

4 sec^{2} (x) - 1

Rewrite

4

2^{2}

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

Rewrite

1

1^{2}

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

Apply exponent rule:

a^{m} b^{m} = (ab)^{m}

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

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

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

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

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

= (2 sec (x) + 1) (2 sec (x) - 1)

= - (2 sec (x) + 1) (2 sec (x) - 1)

= - \frac{( 2 sec ( x ) + 1 ) ( 2 sec ( x ) - 1 )}{1 + 2 sec ( x )}

Cancel the common factor:

2 sec (x) + 1

= - (2 sec (x) - 1)

Negate

- (2 sec (x) - 1) = - 2 sec (x) + 1

= - 2 sec (x) + 1

= 1 - 2 sec (x)

We showed that the two sides could take the same form

\Rightarrow True

p ro v e tan (x) = cot (\frac{π}{2} - x)

p ro v e sec (2 x) = \frac{csc ^{2} ( x )}{csc ^{2} ( x ) - 2}

p ro v e \frac{1 - cos ( a )}{1 + cos ( a )} = tan^{2} (\frac{a}{2})

p ro v e csc^{2} (x) - cos (x) sec (x) = cot^{2} (x)

p ro v e \frac{csc ( x ) + 1}{cot ( x ) + cos ( x )} = sec (x)

Is (1-4sec^2(x))/(1+2sec(x))=1-2sec(x) ?
The answer to whether (1-4sec^2(x))/(1+2sec(x))=1-2sec(x) is True