Is 1/((sec(θ))^2)+1/((csc(θ))^2)=1 ?

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

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prove 1/((sec(θ))^2)+1/((csc(θ))^2)=1

prove

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

True

Solution steps

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

Manipulating left side

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

Simplify

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

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

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

(\frac{1}{sin ( θ )})^{2}

Apply exponent rule:

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

= \frac{1 ^{2}}{sin ^{2} ( θ )}

Apply rule

1^{a} = 1

1^{2} = 1

= \frac{1}{sin ^{2} ( θ )}

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

Apply the fraction rule:

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

= \frac{sin ^{2} ( θ )}{1}

Apply rule

\frac{a}{1} = a

= sin^{2} (θ)

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

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

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

(\frac{1}{cos ( θ )})^{2}

Apply exponent rule:

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

= \frac{1 ^{2}}{cos ^{2} ( θ )}

Apply rule

1^{a} = 1

1^{2} = 1

= \frac{1}{cos ^{2} ( θ )}

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

Apply the fraction rule:

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

= \frac{cos ^{2} ( θ )}{1}

Apply rule

\frac{a}{1} = a

= cos^{2} (θ)

= sin^{2} (θ) + cos^{2} (θ)

= sin^{2} (θ) + cos^{2} (θ)

= cos^{2} (θ) + sin^{2} (θ)

Rewrite using trig identities

cos^{2} (θ) + sin^{2} (θ)

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

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

p ro v e cot (x) + tan (x) = sec (csc (x))

p ro v e cos (\frac{1}{3} π) = cos (- \frac{1}{3} π)

p ro v e (cot (x) + tan (x)) cos (x) = csc (x)

p ro v e cos (- \frac{5 π}{4}) = cos (\frac{3 π}{4})

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