What is the general solution for (csc(b))/(csc(b))= 1/(sqrt(1-sin^2(b))) ?

The general solution for (csc(b))/(csc(b))= 1/(sqrt(1-sin^2(b))) is No Solution for b\in\mathbb{R}

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(csc(b))/(csc(b))= 1/(sqrt(1-sin^2(b)))

Solution

\frac{csc ( b )}{csc ( b )} = \frac{1}{1 - sin ^{2} ( b )}

Solution

No Solution for b \in R

Solution steps

\frac{csc ( b )}{csc ( b )} = \frac{1}{1 - sin ^{2} ( b )}

\frac{csc ( b )}{csc ( b )} = 1

\frac{csc ( b )}{csc ( b )}

Apply rule

\frac{a}{a} = 1

= 1

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

Multiply both sides by

1 - sin^{2} (b)

1 \cdot 1 - sin^{2} (b) = \frac{1}{1 - sin ^{2} ( b )} 1 - sin^{2} (b)

Simplify

1 - sin^{2} (b) = 1

Square both sides:

1 - sin^{2} (b) = 1

1 - sin^{2} (b) = 1

(1 - sin^{2} (b))^{2} = 1^{2}

Expand

(1 - sin^{2} (b))^{2} : 1 - sin^{2} (b)

(1 - sin^{2} (b))^{2}

Apply radical rule:

a = a^{\frac{1}{2}}

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

Apply exponent rule:

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

= (1 - sin^{2} (b))^{\frac{1}{2} \cdot 2}

\frac{1}{2} \cdot 2 = 1

\frac{1}{2} \cdot 2

Multiply fractions:

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

= \frac{1 \cdot 2}{2}

Cancel the common factor:

2

= 1

= 1 - sin^{2} (b)

Expand

1^{2} : 1

1^{2}

Apply rule

1^{a} = 1

= 1

1 - sin^{2} (b) = 1

1 - sin^{2} (b) = 1

Solve

1 - sin^{2} (b) = 1 : sin (b) = 0

1 - sin^{2} (b) = 1

Move

1

to the right side

1 - sin^{2} (b) = 1

Subtract

1

from both sides

1 - sin^{2} (b) - 1 = 1 - 1

Simplify

- sin^{2} (b) = 0

- sin^{2} (b) = 0

Divide both sides by

- 1

- sin^{2} (b) = 0

Divide both sides by

- 1

- sin^{2} (b) = 0

Divide both sides by

- 1

\frac{- sin ^{2} ( b )}{- 1} = \frac{0}{- 1}

Simplify

sin^{2} (b) = 0

sin^{2} (b) = 0

Apply rule

x^{n} = 0 \Rightarrow x = 0

sin (b) = 0

sin (b) = 0

Verify Solutions:

sin (b) = 0

True

Check the solutions by plugging them into

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

Remove the ones that don't agree with the equation.

Plug in

sin (b) = 0 :

True

1 = \frac{1}{1 - 0 ^{2}}

\frac{1}{1 - 0 ^{2}} = 1

\frac{1}{1 - 0 ^{2}}

Apply rule

0^{a} = 0

0^{2} = 0

= \frac{1}{1 - 0}

1 - 0 = 1

1 - 0

Subtract the numbers:

1 - 0 = 1

= 1

Apply rule

1 = 1

= 1

= \frac{1}{1}

Apply rule

\frac{a}{1} = a

= 1

1 = 1

True

The solution is

sin (b) = 0

General solutions for

sin (b) = 0

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

b = 0 + 2 πn, b = π + 2 πn

b = 0 + 2 πn, b = π + 2 πn

Solve

b = 0 + 2 πn : b = 2 πn

b = 0 + 2 πn

0 + 2 πn = 2 πn

b = 2 πn

b = 2 πn, b = π + 2 πn

Since the equation is undefined for:

2 πn, π + 2 πn

No Solution for b \in R

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Frequently Asked Questions (FAQ)

What is the general solution for (csc(b))/(csc(b))= 1/(sqrt(1-sin^2(b))) ?
The general solution for (csc(b))/(csc(b))= 1/(sqrt(1-sin^2(b))) is No Solution for b\in\mathbb{R}