What is the general solution for sin^2(x)(tan^2(x))=1-sin^2(x) ?

The general solution for sin^2(x)(tan^2(x))=1-sin^2(x) is x=(3pi)/4+pin,x= pi/4+pin

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sin^2(x)(tan^2(x))=1-sin^2(x)

Solution

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

Solution

x = \frac{3 π}{4} + πn, x = \frac{π}{4} + πn

Degrees

x = 13 5^{\circ} + 18 0^{\circ} n, x = 4 5^{\circ} + 18 0^{\circ} n

Solution steps

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

Subtract

1 - sin^{2} (x)

from both sides

sin^{2} (x) tan^{2} (x) - 1 + sin^{2} (x) = 0

Rewrite using trig identities

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

Use the Pythagorean identity:

tan^{2} (x) + 1 = sec^{2} (x)

tan^{2} (x) = sec^{2} (x) - 1

= - 1 + sin^{2} (x) + sin^{2} (x) (sec^{2} (x) - 1)

Simplify

- 1 + sin^{2} (x) + sin^{2} (x) (sec^{2} (x) - 1) : sin^{2} (x) sec^{2} (x) - 1

- 1 + sin^{2} (x) + sin^{2} (x) (sec^{2} (x) - 1)

Expand

sin^{2} (x) (sec^{2} (x) - 1) : sin^{2} (x) sec^{2} (x) - sin^{2} (x)

sin^{2} (x) (sec^{2} (x) - 1)

Apply the distributive law:

a (b - c) = ab - a c

a = sin^{2} (x), b = sec^{2} (x), c = 1

= sin^{2} (x) sec^{2} (x) - sin^{2} (x) \cdot 1

= sin^{2} (x) sec^{2} (x) - 1 \cdot sin^{2} (x)

Multiply:

1 \cdot sin^{2} (x) = sin^{2} (x)

= sin^{2} (x) sec^{2} (x) - sin^{2} (x)

= - 1 + sin^{2} (x) + sin^{2} (x) sec^{2} (x) - sin^{2} (x)

Simplify

- 1 + sin^{2} (x) + sin^{2} (x) sec^{2} (x) - sin^{2} (x) : sin^{2} (x) sec^{2} (x) - 1

- 1 + sin^{2} (x) + sin^{2} (x) sec^{2} (x) - sin^{2} (x)

Group like terms

= sin^{2} (x) + sin^{2} (x) sec^{2} (x) - sin^{2} (x) - 1

Add similar elements:

sin^{2} (x) - sin^{2} (x) = 0

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

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

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

- 1 + sec^{2} (x) sin^{2} (x) = 0

Factor

- 1 + sec^{2} (x) sin^{2} (x) : (sec (x) sin (x) + 1) (sec (x) sin (x) - 1)

- 1 + sec^{2} (x) sin^{2} (x)

Apply exponent rule:

a^{m} b^{m} = (ab)^{m}

sec^{2} (x) sin^{2} (x) = (sec (x) sin (x))^{2}

= - 1 + (sec (x) sin (x))^{2}

Apply Difference of Two Squares Formula:

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

- 1 + (sec (x) sin (x))^{2} = (sec (x) sin (x) + 1) (sec (x) sin (x) - 1)

= (sec (x) sin (x) + 1) (sec (x) sin (x) - 1)

(sec (x) sin (x) + 1) (sec (x) sin (x) - 1) = 0

Solving each part separately

sec (x) sin (x) + 1 = 0 or sec (x) sin (x) - 1 = 0

sec (x) sin (x) + 1 = 0 : x = \frac{3 π}{4} + πn

sec (x) sin (x) + 1 = 0

Rewrite using trig identities

sec (x) sin (x) + 1

sec (x) sin (x) = tan (x)

sec (x) sin (x)

Express with sin, cos

sec (x) sin (x)

Use the basic trigonometric identity:

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

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

Simplify

\frac{1}{cos ( x )} sin (x) : \frac{sin ( x )}{cos ( x )}

\frac{1}{cos ( x )} sin (x)

Multiply fractions:

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

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

Multiply:

1 \cdot sin (x) = sin (x)

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

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

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

Use the basic trigonometric identity:

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

= tan (x)

= 1 + tan (x)

1 + tan (x) = 0

Move

1

to the right side

1 + tan (x) = 0

Subtract

1

from both sides

1 + tan (x) - 1 = 0 - 1

Simplify

tan (x) = - 1

tan (x) = - 1

General solutions for

tan (x) = - 1

tan (x)

periodicity table with

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

x = \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

sec (x) sin (x) - 1 = 0 : x = \frac{π}{4} + πn

sec (x) sin (x) - 1 = 0

Rewrite using trig identities

sec (x) sin (x) - 1

sec (x) sin (x) = tan (x)

sec (x) sin (x)

Express with sin, cos

sec (x) sin (x)

Use the basic trigonometric identity:

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

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

Simplify

\frac{1}{cos ( x )} sin (x) : \frac{sin ( x )}{cos ( x )}

\frac{1}{cos ( x )} sin (x)

Multiply fractions:

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

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

Multiply:

1 \cdot sin (x) = sin (x)

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

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

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

Use the basic trigonometric identity:

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

= tan (x)

= - 1 + tan (x)

- 1 + tan (x) = 0

Move

1

to the right side

- 1 + tan (x) = 0

Add

1

to both sides

- 1 + tan (x) + 1 = 0 + 1

Simplify

tan (x) = 1

tan (x) = 1

General solutions for

tan (x) = 1

tan (x)

periodicity table with

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

x = \frac{π}{4} + πn

x = \frac{π}{4} + πn

Combine all the solutions

x = \frac{3 π}{4} + πn, x = \frac{π}{4} + πn

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Frequently Asked Questions (FAQ)

What is the general solution for sin^2(x)(tan^2(x))=1-sin^2(x) ?
The general solution for sin^2(x)(tan^2(x))=1-sin^2(x) is x=(3pi)/4+pin,x= pi/4+pin