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

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

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prove (1+tan^2(θ))/(csc(θ))=tan(θ)sec(θ)

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

\frac{1 + tan ^{2} ( θ )}{csc ( θ )}

Express with sin, cos

\frac{1 + tan ^{2} ( θ )}{csc ( θ )}

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply the fraction rule:

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

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

Apply exponent rule:

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

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

Join

1 + \frac{sin ^{2} ( θ )}{cos ^{2} ( θ )} : \frac{cos ^{2} ( θ ) + sin ^{2} ( θ )}{cos ^{2} ( θ )}

1 + \frac{sin ^{2} ( θ )}{cos ^{2} ( θ )}

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

Multiply:

1 \cdot cos^{2} (θ) = cos^{2} (θ)

= \frac{cos ^{2} ( θ ) + sin ^{2} ( θ )}{cos ^{2} ( θ )}

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

Apply the fraction rule:

\frac{a}{1} = a

= \frac{cos ^{2} ( θ ) + sin ^{2} ( θ )}{cos ^{2} ( θ )} sin (θ)

Multiply fractions:

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

= \frac{( cos ^{2} ( θ ) + sin ^{2} ( θ ) ) sin ( θ )}{cos ^{2} ( θ )}

= \frac{sin ( θ ) ( cos ^{2} ( θ ) + sin ^{2} ( θ ) )}{cos ^{2} ( θ )}

= \frac{( cos ^{2} ( θ ) + sin ^{2} ( θ ) ) sin ( θ )}{cos ^{2} ( θ )}

Rewrite using trig identities

\frac{( cos ^{2} ( θ ) + sin ^{2} ( θ ) ) sin ( θ )}{cos ^{2} ( θ )}

Use the Pythagorean identity:

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

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

Simplify

= \frac{sin ( θ )}{cos ^{2} ( θ )}

= \frac{sin ( θ )}{cos ^{2} ( θ )}

Manipulating right side

tan (θ) sec (θ)

Express with sin, cos

sec (θ) tan (θ)

Use the basic trigonometric identity:

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

= \frac{1}{cos ( θ )} tan (θ)

Use the basic trigonometric identity:

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

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

Simplify

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

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

Multiply fractions:

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

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

Multiply:

1 \cdot sin (θ) = sin (θ)

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

cos (θ) cos (θ) = cos^{2} (θ)

cos (θ) cos (θ)

Apply exponent rule:

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

cos (θ) cos (θ) = cos^{1 + 1} (θ)

= cos^{1 + 1} (θ)

Add the numbers:

1 + 1 = 2

= cos^{2} (θ)

= \frac{sin ( θ )}{cos ^{2} ( θ )}

= \frac{sin ( θ )}{cos ^{2} ( θ )}

= \frac{sin ( θ )}{cos ^{2} ( θ )}

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

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