Is csc(x)tan(x)sec(x)=sec^2(x) ?

The answer to whether csc(x)tan(x)sec(x)=sec^2(x) is True

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prove csc(x)tan(x)sec(x)=sec^2(x)

prove

csc (x) tan (x) sec (x) = sec^{2} (x)

True

Solution steps

csc (x) tan (x) sec (x) = sec^{2} (x)

Manipulating left side

csc (x) tan (x) sec (x)

Express with sin, cos

csc (x) sec (x) tan (x)

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

\frac{1}{sin ( x )} \cdot \frac{1}{cos ( x )} \cdot \frac{sin ( x )}{cos ( x )} : \frac{1}{cos ^{2} ( x )}

\frac{1}{sin ( x )} \cdot \frac{1}{cos ( x )} \cdot \frac{sin ( x )}{cos ( x )}

Multiply fractions:

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

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

Cancel the common factor:

sin (x)

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

Multiply the numbers:

1 \cdot 1 = 1

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

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

cos (x) cos (x)

Apply exponent rule:

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

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

= cos^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= cos^{2} (x)

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

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

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

Rewrite using trig identities

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

Apply exponent rule:

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

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

Apply the fraction rule:

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

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

Apply rule

\frac{a}{1} = a

= sec^{2} (x)

sec^{2} (x)

sec^{2} (x)

We showed that the two sides could take the same form

\Rightarrow True

tan (x) sec^{4} (x) = \frac{sin ( x )}{cos ^{5} ( x )}

csc (x) + cot (x) sec (x) - 1 = tan (x)

\frac{3 csc ( x ) - 3 sin ( x )}{tan ( x ) - cot ( x )} = 3 cos^{3} (x)

9 cos (x^{\circ}) + 6 sin (x^{\circ}) = 10

\frac{tan ^{2} ( A )}{sec ^{2} ( A )} = sin^{2} (A)

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