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

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

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prove 1/(sec^2(x))+1/(csc^2(x))=1

prove

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

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

csc (x) = \frac{1}{sin ( x )}

= \frac{1}{( \frac{1}{s i n ( x )} ) ^{2}} + \frac{1}{sec ^{2} ( x )}

Use the basic trigonometric identity:

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

= \frac{1}{( \frac{1}{s i n ( x )} ) ^{2}} + \frac{1}{( \frac{1}{c o s ( x )} ) ^{2}}

Simplify

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

\frac{1}{( \frac{1}{s i n ( x )} ) ^{2}} + \frac{1}{( \frac{1}{c o s ( x )} ) ^{2}}

\frac{1}{( \frac{1}{s i n ( x )} ) ^{2}} = sin^{2} (x)

\frac{1}{( \frac{1}{s i n ( x )} ) ^{2}}

(\frac{1}{sin ( x )})^{2} = \frac{1}{sin ^{2} ( x )}

(\frac{1}{sin ( x )})^{2}

Apply exponent rule:

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

= \frac{1 ^{2}}{sin ^{2} ( x )}

Apply rule

1^{a} = 1

1^{2} = 1

= \frac{1}{sin ^{2} ( x )}

= \frac{1}{\frac{1}{s i n ^{2} ( x )}}

Apply the fraction rule:

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

= \frac{sin ^{2} ( x )}{1}

Apply rule

\frac{a}{1} = a

= sin^{2} (x)

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

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

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

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

Apply exponent rule:

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

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

Apply the fraction rule:

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

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

Apply rule

\frac{a}{1} = a

= cos^{2} (x)

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

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

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

Rewrite using trig identities

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

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

(1 + tan^{2} (x)) cot^{2} (x) = csc^{2} (x)

csc (θ) sec (θ) - cot (θ) = tan (θ)

1 - tan^{2} (\frac{θ}{2}) = \frac{2 cos ( θ )}{1 + cos ( θ )}

sin (x + y) sec (x) sec (y) = tan (x) + tan (y)

\frac{sin ( x ) - csc ( x )}{csc ( x )} = - cos^{2} (x)

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