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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

a = 2 πn, a = π + 2 πn

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 ) - 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 (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 : a = 2 πn, a = π + 2 πn

Combine all the solutions

a = 2 πn, a = π + 2 πn

tan (a) = \frac{5}{9}, b = 6

sin^{3} (x) sin^{2} (x) sin (x) = 0

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

sin^{2} (x) cos^{2} (x) = \frac{( 1 - cos ^{4} ( x ) )}{8}

sin^{3} (x) + sin (x) = 2 sin^{22} (x)

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