Is tan(β)=csc(β)sec(β)-cot(β) ?

The answer to whether tan(β)=csc(β)sec(β)-cot(β) is True

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

prove tan(β)=csc(β)sec(β)-cot(β)

Solution

prove

tan (β) = csc (β) sec (β) - cot (β)

Solution

True

Solution steps

tan (β) = csc (β) sec (β) - cot (β)

Manipulating right side

csc (β) sec (β) - cot (β)

Express with sin, cos

- cot (β) + csc (β) sec (β)

Use the basic trigonometric identity:

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

= - \frac{cos ( β )}{sin ( β )} + csc (β) sec (β)

Use the basic trigonometric identity:

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

= - \frac{cos ( β )}{sin ( β )} + \frac{1}{sin ( β )} sec (β)

Use the basic trigonometric identity:

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

= - \frac{cos ( β )}{sin ( β )} + \frac{1}{sin ( β )} \cdot \frac{1}{cos ( β )}

Simplify

- \frac{cos ( β )}{sin ( β )} + \frac{1}{sin ( β )} \cdot \frac{1}{cos ( β )} : \frac{- cos ^{2} ( β ) + 1}{sin ( β ) cos ( β )}

- \frac{cos ( β )}{sin ( β )} + \frac{1}{sin ( β )} \cdot \frac{1}{cos ( β )}

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

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

Multiply fractions:

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

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

Multiply the numbers:

1 \cdot 1 = 1

= \frac{1}{sin ( β ) cos ( β )}

= - \frac{cos ( β )}{sin ( β )} + \frac{1}{sin ( β ) cos ( β )}

Least Common Multiplier of

sin (β), sin (β) cos (β) : sin (β) cos (β)

sin (β), sin (β) cos (β)

Lowest Common Multiplier (LCM)

Compute an expression comprised of factors that appear either in

sin (β)

sin (β) cos (β)

= sin (β) cos (β)

Adjust Fractions based on the LCM

Multiply each numerator by the same amount needed to multiply its

corresponding denominator to turn it into the LCM

sin (β) cos (β)

For

\frac{cos ( β )}{sin ( β )} :

multiply the denominator and numerator by

cos (β)

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

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

Since the denominators are equal, combine the fractions:

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Cancel the common factor:

sin (β)

= \frac{sin ( β )}{cos ( β )}

= \frac{sin ( β )}{cos ( β )}

Use the basic trigonometric identity:

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

= tan (β)

We showed that the two sides could take the same form

\Rightarrow True

Popular Examples

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

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

prove sec^2(θ)(1-sin^2(θ))=sec^2(θ)prove

sec^{2} (θ) (1 - sin^{2} (θ)) = sec^{2} (θ)

prove cos(x-90)=sin(x)prove

cos (x - 9 0^{\circ}) = sin (x)

prove (1+sec(B))/(tan(B)+sin(B))=csc(B)prove

\frac{1 + sec ( B )}{tan ( B ) + sin ( B )} = csc (B)

prove (sin(θ))/(cos^2(θ))=tan(θ)sec(θ)prove

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

Frequently Asked Questions (FAQ)

Is tan(β)=csc(β)sec(β)-cot(β) ?
The answer to whether tan(β)=csc(β)sec(β)-cot(β) is True