What is the general solution for (sin^2(a)+1)/((1+tan^2(a)))=1 ?

The general solution for (sin^2(a)+1)/((1+tan^2(a)))=1 is a=2pin,a=pi+2pin

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Popular >

(sin^2(a)+1)/((1+tan^2(a)))=1

Solution

\frac{sin ^{2} ( a ) + 1}{( 1 + tan ^{2} ( a ) )} = 1

Solution

a = 2 πn, a = π + 2 πn

Degrees

a = 0^{\circ} + 36 0^{\circ} n, a = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

\frac{sin ^{2} ( a ) + 1}{( 1 + tan ^{2} ( a ) )} = 1

Subtract

1

from both sides

\frac{sin ^{2} ( a ) + 1}{1 + tan ^{2} ( a )} - 1 = 0

Simplify

\frac{sin ^{2} ( a ) + 1}{1 + tan ^{2} ( a )} - 1 : \frac{sin ^{2} ( a ) - tan ^{2} ( a )}{1 + tan ^{2} ( a )}

\frac{sin ^{2} ( a ) + 1}{1 + tan ^{2} ( a )} - 1

Convert element to fraction:

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

= \frac{sin ^{2} ( a ) + 1}{1 + tan ^{2} ( a )} - \frac{1 \cdot ( 1 + tan ^{2} ( a ) )}{1 + tan ^{2} ( a )}

Since the denominators are equal, combine the fractions:

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

= \frac{sin ^{2} ( a ) + 1 - 1 \cdot ( 1 + tan ^{2} ( a ) )}{1 + tan ^{2} ( a )}

Multiply:

1 \cdot (1 + tan^{2} (a)) = (1 + tan^{2} (a))

= \frac{sin ^{2} ( a ) + 1 - ( tan ^{2} ( a ) + 1 )}{1 + tan ^{2} ( a )}

Expand

sin^{2} (a) + 1 - (1 + tan^{2} (a)) : sin^{2} (a) - tan^{2} (a)

sin^{2} (a) + 1 - (1 + tan^{2} (a))

- (1 + tan^{2} (a)) : - 1 - tan^{2} (a)

- (1 + tan^{2} (a))

Distribute parentheses

= - (1) - (tan^{2} (a))

Apply minus-plus rules

+ (- a) = - a

= - 1 - tan^{2} (a)

= sin^{2} (a) + 1 - 1 - tan^{2} (a)

1 - 1 = 0

= sin^{2} (a) - tan^{2} (a)

= \frac{sin ^{2} ( a ) - tan ^{2} ( a )}{1 + tan ^{2} ( a )}

\frac{sin ^{2} ( a ) - tan ^{2} ( a )}{1 + tan ^{2} ( a )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

sin^{2} (a) - tan^{2} (a) = 0

Factor

sin^{2} (a) - tan^{2} (a) : (sin (a) + tan (a)) (sin (a) - tan (a))

sin^{2} (a) - tan^{2} (a)

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

sin^{2} (a) - tan^{2} (a) = (sin (a) + tan (a)) (sin (a) - tan (a))

= (sin (a) + tan (a)) (sin (a) - tan (a))

(sin (a) + tan (a)) (sin (a) - tan (a)) = 0

Solving each part separately

sin (a) + tan (a) = 0 or sin (a) - tan (a) = 0

sin (a) + tan (a) = 0 : a = 2 πn, a = π + 2 πn

sin (a) + tan (a) = 0

Express with sin, cos

sin (a) + tan (a)

Use the basic trigonometric identity:

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

= sin (a) + \frac{sin ( a )}{cos ( a )}

Simplify

sin (a) + \frac{sin ( a )}{cos ( a )} : \frac{sin ( a ) cos ( a ) + sin ( a )}{cos ( a )}

sin (a) + \frac{sin ( a )}{cos ( a )}

Convert element to fraction:

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

= \frac{sin ( a ) cos ( a )}{cos ( a )} + \frac{sin ( a )}{cos ( a )}

Since the denominators are equal, combine the fractions:

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

= \frac{sin ( a ) cos ( a ) + sin ( a )}{cos ( a )}

= \frac{sin ( a ) cos ( a ) + sin ( a )}{cos ( a )}

\frac{sin ( a ) + cos ( a ) sin ( a )}{cos ( a )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

sin (a) + cos (a) sin (a) = 0

Factor

sin (a) + cos (a) sin (a) : sin (a) (cos (a) + 1)

sin (a) + cos (a) sin (a)

Factor out common term

sin (a)

= sin (a) (1 + cos (a))

sin (a) (cos (a) + 1) = 0

Solving each part separately

sin (a) = 0 or cos (a) + 1 = 0

sin (a) = 0 : a = 2 πn, a = π + 2 πn

sin (a) = 0

General solutions for

sin (a) = 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}

a = 0 + 2 πn, a = π + 2 πn

a = 0 + 2 πn, a = π + 2 πn

Solve

a = 0 + 2 πn : a = 2 πn

a = 0 + 2 πn

0 + 2 πn = 2 πn

a = 2 πn

a = 2 πn, a = π + 2 πn

cos (a) + 1 = 0 : a = π + 2 πn

cos (a) + 1 = 0

Move

1

to the right side

cos (a) + 1 = 0

Subtract

1

from both sides

cos (a) + 1 - 1 = 0 - 1

Simplify

cos (a) = - 1

cos (a) = - 1

General solutions for

cos (a) = - 1

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

a = π + 2 πn

a = π + 2 πn

Combine all the solutions

a = 2 πn, a = π + 2 πn

sin (a) - tan (a) = 0 : a = 2 πn, a = π + 2 πn

sin (a) - tan (a) = 0

Express with sin, cos

sin (a) - tan (a)

Use the basic trigonometric identity:

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

= sin (a) - \frac{sin ( a )}{cos ( a )}

Simplify

sin (a) - \frac{sin ( a )}{cos ( a )} : \frac{sin ( a ) cos ( a ) - sin ( a )}{cos ( a )}

sin (a) - \frac{sin ( a )}{cos ( a )}

Convert element to fraction:

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

= \frac{sin ( a ) cos ( a )}{cos ( a )} - \frac{sin ( a )}{cos ( a )}

Since the denominators are equal, combine the fractions:

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

= \frac{sin ( a ) cos ( a ) - sin ( a )}{cos ( a )}

= \frac{sin ( a ) cos ( a ) - sin ( a )}{cos ( a )}

\frac{- sin ( a ) + cos ( a ) sin ( a )}{cos ( a )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

- sin (a) + cos (a) sin (a) = 0

Factor

- sin (a) + cos (a) sin (a) : sin (a) (cos (a) - 1)

- sin (a) + cos (a) sin (a)

Factor out common term

sin (a)

= sin (a) (- 1 + cos (a))

sin (a) (cos (a) - 1) = 0

Solving each part separately

sin (a) = 0 or cos (a) - 1 = 0

sin (a) = 0 : a = 2 πn, a = π + 2 πn

sin (a) = 0

General solutions for

sin (a) = 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}

a = 0 + 2 πn, a = π + 2 πn

a = 0 + 2 πn, a = π + 2 πn

Solve

a = 0 + 2 πn : a = 2 πn

a = 0 + 2 πn

0 + 2 πn = 2 πn

a = 2 πn

a = 2 πn, a = π + 2 πn

cos (a) - 1 = 0 : a = 2 πn

cos (a) - 1 = 0

Move

1

to the right side

cos (a) - 1 = 0

Add

1

to both sides

cos (a) - 1 + 1 = 0 + 1

Simplify

cos (a) = 1

cos (a) = 1

General solutions for

cos (a) = 1

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

a = 0 + 2 πn

a = 0 + 2 πn

Solve

a = 0 + 2 πn : a = 2 πn

a = 0 + 2 πn

0 + 2 πn = 2 πn

a = 2 πn

a = 2 πn

Combine all the solutions

a = 2 πn, a = π + 2 πn

Combine all the solutions

a = 2 πn, a = π + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for (sin^2(a)+1)/((1+tan^2(a)))=1 ?
The general solution for (sin^2(a)+1)/((1+tan^2(a)))=1 is a=2pin,a=pi+2pin