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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

prove

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

True

Solution steps

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

Manipulating left side

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

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

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

Factor

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

sec^{2} (x) - 1

Rewrite

1

1^{2}

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

Apply Difference of Two Squares Formula:

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

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

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

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

Cancel the common factor:

sec (x) + 1

= sec (x) - 1

= sec (x) - 1

= sec (x) - 1

We showed that the two sides could take the same form

\Rightarrow True

(1 - cos (θ)) (1 + cos (θ)) = sin^{2} (θ)

\frac{1 - 3 cos ( x ) - 4 cos ^{2} ( x )}{sin ^{2} ( x )} = \frac{1 - 4 cos ( x )}{1 - cos ( x )}

(1 + cos (2 x)) (1 - cos (2 x)) = sin^{2} (2 x)

\frac{sin ( x ) csc ( x )}{cot ( x )} = tan (x)

cos^{2} (θ) (tan^{2} (θ) + 1) = 1

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