Is csc(2θ)=(1+tan^2(θ))/(2tan(θ)) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

prove csc(2θ)=(1+tan^2(θ))/(2tan(θ))

Solution

prove

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

Solution

True

Solution steps

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

Manipulating right side

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

Express with sin, cos

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

Simplify

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

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

Multiply

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

2 \cdot \frac{sin ( θ )}{cos ( θ )}

Multiply fractions:

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

= \frac{sin ( θ ) \cdot 2}{cos ( θ )}

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

Apply exponent rule:

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

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

Apply the fraction rule:

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

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

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 )} cos ( θ )}{2 sin ( θ )}

Multiply

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos (θ)

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

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

Apply the fraction rule:

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

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

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

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

Rewrite using trig identities

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

Use the Double Angle identity:

2 sin (x) cos (x) = sin (2 x)

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

Use the Pythagorean identity:

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

= \frac{1}{sin ( 2 θ )}

= \frac{1}{sin ( 2 θ )}

Rewrite using trig identities

Use the basic trigonometric identity:

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

\frac{1}{\frac{1}{c s c ( 2 θ )}}

Simplify

\frac{1}{\frac{1}{c s c ( 2 θ )}}

Apply the fraction rule:

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

= \frac{csc ( 2 θ )}{1}

Apply rule

\frac{a}{1} = a

= csc (2 θ)

csc (2 θ)

csc (2 θ)

We showed that the two sides could take the same form

\Rightarrow True

Popular Examples

prove tan^2(x)(1+cos(2x))=1-cos(2x)prove

tan^{2} (x) (1 + cos (2 x)) = 1 - cos (2 x)

prove (1+tan(θ))/(sin(θ)+cos(θ))=sec(θ)prove

\frac{1 + tan ( θ )}{sin ( θ ) + cos ( θ )} = sec (θ)

prove (cos(x)sec(x))/(cot(x))=tan(x)prove

\frac{cos ( x ) sec ( x )}{cot ( x )} = tan (x)

prove 8cos^4(θ-8)sin^4(θ)=8cos(2θ)prove

8 cos^{4} (θ - 8) sin^{4} (θ) = 8 cos (2 θ)

prove csc(θ)sec(θ)-csc(θ)cos(θ)=tan(θ)prove

csc (θ) sec (θ) - csc (θ) cos (θ) = tan (θ)

Frequently Asked Questions (FAQ)

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