Is (sec(t))/(cos(t))-(tan(t))/(cot(t))=1 ?

The answer to whether (sec(t))/(cos(t))-(tan(t))/(cot(t))=1 is True

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prove (sec(t))/(cos(t))-(tan(t))/(cot(t))=1

prove

\frac{sec ( t )}{cos ( t )} - \frac{tan ( t )}{cot ( t )} = 1

True

Solution steps

\frac{sec ( t )}{cos ( t )} - \frac{tan ( t )}{cot ( t )} = 1

Manipulating left side

\frac{sec ( t )}{cos ( t )} - \frac{tan ( t )}{cot ( t )}

Express with sin, cos

\frac{sec ( t )}{cos ( t )} - \frac{tan ( t )}{cot ( t )}

Use the basic trigonometric identity:

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

= \frac{\frac{1}{c o s ( t )}}{cos ( t )} - \frac{tan ( t )}{cot ( t )}

Use the basic trigonometric identity:

tan (x) = \frac{sin ( x )}{cos ( x )}

= \frac{\frac{1}{c o s ( t )}}{cos ( t )} - \frac{\frac{s i n ( t )}{c o s ( t )}}{cot ( t )}

Use the basic trigonometric identity:

cot (x) = \frac{cos ( x )}{sin ( x )}

= \frac{\frac{1}{c o s ( t )}}{cos ( t )} - \frac{\frac{s i n ( t )}{c o s ( t )}}{\frac{c o s ( t )}{s i n ( t )}}

Simplify

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

\frac{\frac{1}{c o s ( t )}}{cos ( t )} - \frac{\frac{s i n ( t )}{c o s ( t )}}{\frac{c o s ( t )}{s i n ( t )}}

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

\frac{\frac{1}{c o s ( t )}}{cos ( t )}

Apply the fraction rule:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

= \frac{1}{cos ( t ) cos ( t )}

cos (t) cos (t) = cos^{2} (t)

cos (t) cos (t)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

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

= cos^{1 + 1} (t)

Add the numbers:

1 + 1 = 2

= cos^{2} (t)

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

\frac{\frac{s i n ( t )}{c o s ( t )}}{\frac{c o s ( t )}{s i n ( t )}} = \frac{sin ^{2} ( t )}{cos ^{2} ( t )}

\frac{\frac{s i n ( t )}{c o s ( t )}}{\frac{c o s ( t )}{s i n ( t )}}

Divide fractions:

\frac{\frac{a}{b}}{\frac{c}{d}} = \frac{a \cdot d}{b \cdot c}

= \frac{sin ( t ) sin ( t )}{cos ( t ) cos ( t )}

sin (t) sin (t) = sin^{2} (t)

sin (t) sin (t)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

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

= sin^{1 + 1} (t)

Add the numbers:

1 + 1 = 2

= sin^{2} (t)

= \frac{sin ^{2} ( t )}{cos ( t ) cos ( t )}

cos (t) cos (t) = cos^{2} (t)

cos (t) cos (t)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

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

= cos^{1 + 1} (t)

Add the numbers:

1 + 1 = 2

= cos^{2} (t)

= \frac{sin ^{2} ( t )}{cos ^{2} ( t )}

= \frac{1}{cos ^{2} ( t )} - \frac{sin ^{2} ( t )}{cos ^{2} ( t )}

Apply rule

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{1 - sin ^{2} ( t )}{cos ^{2} ( t )}

= \frac{1 - sin ^{2} ( t )}{cos ^{2} ( t )}

= \frac{1 - sin ^{2} ( t )}{cos ^{2} ( t )}

Rewrite using trig identities

\frac{1 - sin ^{2} ( t )}{cos ^{2} ( t )}

Use the Pythagorean identity:

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

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

= \frac{cos ^{2} ( t )}{cos ^{2} ( t )}

Apply rule

\frac{a}{a} = 1

= 1

= 1

We showed that the two sides could take the same form

\Rightarrow True

\frac{cot ( y )}{sec ( y )} = csc (y) - sin (y)

\frac{1}{cot ( x )} = tan (x)

csc (θ) = cos (θ) cot (θ) + sin (θ)

cos^{4} (x) + sin^{2} (x) cos^{2} (x) = cos^{2} (x)

\frac{tan ( x ) cos ( x )}{sin ( x )} = 1

Is (sec(t))/(cos(t))-(tan(t))/(cot(t))=1 ?
The answer to whether (sec(t))/(cos(t))-(tan(t))/(cot(t))=1 is True