Is sin^2(v)+1/(sec^2(v))=1 ?

The answer to whether sin^2(v)+1/(sec^2(v))=1 is True

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

prove sin^2(v)+1/(sec^2(v))=1

prove

sin^{2} (v) + \frac{1}{sec ^{2} ( v )} = 1

True

Solution steps

sin^{2} (v) + \frac{1}{sec ^{2} ( v )} = 1

Manipulating left side

sin^{2} (v) + \frac{1}{sec ^{2} ( v )}

Express with sin, cos

\frac{1}{sec ^{2} ( v )} + sin^{2} (v)

Use the basic trigonometric identity:

sec (x) = \frac{1}{cos ( x )}

= \frac{1}{( \frac{1}{c o s ( v )} ) ^{2}} + sin^{2} (v)

Simplify

\frac{1}{( \frac{1}{c o s ( v )} ) ^{2}} + sin^{2} (v) : cos^{2} (v) + sin^{2} (v)

\frac{1}{( \frac{1}{c o s ( v )} ) ^{2}} + sin^{2} (v)

\frac{1}{( \frac{1}{c o s ( v )} ) ^{2}} = cos^{2} (v)

\frac{1}{( \frac{1}{c o s ( v )} ) ^{2}}

(\frac{1}{cos ( v )})^{2} = \frac{1}{cos ^{2} ( v )}

(\frac{1}{cos ( v )})^{2}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{1 ^{2}}{cos ^{2} ( v )}

Apply rule

1^{a} = 1

1^{2} = 1

= \frac{1}{cos ^{2} ( v )}

= \frac{1}{\frac{1}{c o s ^{2} ( v )}}

Apply the fraction rule:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

= \frac{cos ^{2} ( v )}{1}

Apply rule

\frac{a}{1} = a

= cos^{2} (v)

= cos^{2} (v) + sin^{2} (v)

= cos^{2} (v) + sin^{2} (v)

= cos^{2} (v) + sin^{2} (v)

Rewrite using trig identities

cos^{2} (v) + sin^{2} (v)

Use the Pythagorean identity:

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

= 1

= 1

We showed that the two sides could take the same form

\Rightarrow True

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

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

\frac{sin ( 9 θ ) + sin ( 3 θ )}{cos ( 9 θ ) + cos ( 3 θ )} = tan (6 θ)

cot^{2} (A) - cos^{2} (A) = cot^{2} (A) cos^{2} (A)

cos^{2} (2 t) + sin^{2} (2 t) = 1

Is sin^2(v)+1/(sec^2(v))=1 ?
The answer to whether sin^2(v)+1/(sec^2(v))=1 is True