What is the general solution for (sec^2(a))cos^2(a)=sin^2(a) ?

The general solution for (sec^2(a))cos^2(a)=sin^2(a) is No Solution for a\in\mathbb{R}

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(sec^2(a))cos^2(a)=sin^2(a)

Solution

(sec^{2} (a)) cos^{2} (a) = sin^{2} (a)

Solution

No Solution for a \in R

Solution steps

(sec^{2} (a)) cos^{2} (a) = sin^{2} (a)

Subtract

sin^{2} (a)

from both sides

sec^{2} (a) cos^{2} (a) - sin^{2} (a) = 0

Factor

sec^{2} (a) cos^{2} (a) - sin^{2} (a) : (sec (a) cos (a) + sin (a)) (sec (a) cos (a) - sin (a))

sec^{2} (a) cos^{2} (a) - sin^{2} (a)

Rewrite

sec^{2} (a) cos^{2} (a)

(sec (a) cos (a))^{2}

sec^{2} (a) cos^{2} (a)

Apply exponent rule:

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

sec^{2} (a) cos^{2} (a) = (sec (a) cos (a))^{2}

= (sec (a) cos (a))^{2}

= (sec (a) cos (a))^{2} - sin^{2} (a)

Apply Difference of Two Squares Formula:

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

(sec (a) cos (a))^{2} - sin^{2} (a) = (sec (a) cos (a) + sin (a)) (sec (a) cos (a) - sin (a))

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

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

Rewrite using trig identities

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

sec (a) cos (a) = 1

sec (a) cos (a)

Express with sin, cos

sec (a) cos (a)

Use the basic trigonometric identity:

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

= \frac{1}{cos ( a )} cos (a)

\frac{1}{cos ( a )} cos (a) = 1

\frac{1}{cos ( a )} cos (a)

Multiply fractions:

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

= \frac{1 cos ( a )}{cos ( a )}

Cancel the common factor:

cos (a)

= 1

= 1

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

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

Solving each part separately

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

- sin (a) + 1 = 0 : a = \frac{π}{2} + 2 πn

- sin (a) + 1 = 0

Move

1

to the right side

- sin (a) + 1 = 0

Subtract

1

from both sides

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

Simplify

- sin (a) = - 1

- sin (a) = - 1

Divide both sides by

- 1

- sin (a) = - 1

Divide both sides by

- 1

\frac{- sin ( a )}{- 1} = \frac{- 1}{- 1}

Simplify

sin (a) = 1

sin (a) = 1

General solutions for

sin (a) = 1

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 = \frac{π}{2} + 2 πn

a = \frac{π}{2} + 2 πn

sin (a) + 1 = 0 : a = \frac{3 π}{2} + 2 πn

sin (a) + 1 = 0

Move

1

to the right side

sin (a) + 1 = 0

Subtract

1

from both sides

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

Simplify

sin (a) = - 1

sin (a) = - 1

General solutions for

sin (a) = - 1

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 = \frac{3 π}{2} + 2 πn

a = \frac{3 π}{2} + 2 πn

Combine all the solutions

a = \frac{π}{2} + 2 πn, a = \frac{3 π}{2} + 2 πn

Since the equation is undefined for:

\frac{π}{2} + 2 πn, \frac{3 π}{2} + 2 πn

No Solution for a \in R

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Frequently Asked Questions (FAQ)

What is the general solution for (sec^2(a))cos^2(a)=sin^2(a) ?
The general solution for (sec^2(a))cos^2(a)=sin^2(a) is No Solution for a\in\mathbb{R}