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verificar 1+tan^2(pi/2-x)= 1/(cos^2(pi/2-x))

Solución

verificar

1 + tan^{2} (\frac{π}{2} - x) = \frac{1}{cos ^{2} ( \frac{π}{2} - x )}

Verdadero

Pasos de solución

1 + tan^{2} (\frac{π}{2} - x) = \frac{1}{cos ^{2} ( \frac{π}{2} - x )}

Manipular el lado derecho

1 + tan^{2} (\frac{π}{2} - x)

Re-escribir usando identidades trigonométricas

= 1 + (\frac{cos ( x )}{sin ( x )})^{2}

Aplicar las leyes de los exponentes:

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

= 1 + \frac{cos ^{2} ( x )}{sin ^{2} ( x )}

Simplificar

1 + \frac{cos ^{2} ( x )}{sin ^{2} ( x )} : \frac{sin ^{2} ( x ) + cos ^{2} ( x )}{sin ^{2} ( x )}

= \frac{cos ^{2} ( x ) + sin ^{2} ( x )}{sin ^{2} ( x )}

Re-escribir usando identidades trigonométricas

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

Manipular el lado izquierdo

\frac{1}{cos ^{2} ( \frac{π}{2} - x )}

Re-escribir usando identidades trigonométricas

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

Quitar los parentesis:

(a) = a

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

Se demostró que ambos lados pueden tomar la misma forma

\Rightarrow Verdadero

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

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

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

cos (θ) - sin (θ) sin (2 θ) = cos (θ) cos (2 θ)

1 + (cot (A))^{2} = \cdot (csc (A))^{2}