Respuesta :
The identities used to simplify the expression are the following:
The difference of squares formula, [tex]x^2-y^2=(x-y)(x+y)[/tex].
The well-known Pythagorean trig. identity, [tex]\sin^2\theta+\cos^2\theta=1[/tex], for any angle theta.
Last, the identity [tex]\displaystyle{ \frac{\sin\theta}{\cos\theta}=\tan\theta.[/tex]
Thus, from the difference of squares identity we have
[tex]\displaystyle{ \frac{(1-\cos\theta)(1+\cos\theta)}{(1-\sin\theta)(1+\sin\theta)}= \frac{1-\cos^2\theta}{1-\sin^2\theta}.[/tex]
From the identity [tex]\sin^2\theta+\cos^2\theta=1[/tex], we have
[tex]\sin^2\theta=1-\cos^2\theta[/tex], and [tex]\cos^2\theta=1-\sin^2\theta[/tex], thus
[tex]\displaystyle{ \frac{1-\cos^2\theta}{1-\sin^2\theta}= \frac{\sin^2\theta}{\cos^2\theta}= (\frac{\sin\theta}{\cos\theta})^2=\tan^2\theta.[/tex]
Answer: [tex]\tan^2\theta.[/tex]
The difference of squares formula, [tex]x^2-y^2=(x-y)(x+y)[/tex].
The well-known Pythagorean trig. identity, [tex]\sin^2\theta+\cos^2\theta=1[/tex], for any angle theta.
Last, the identity [tex]\displaystyle{ \frac{\sin\theta}{\cos\theta}=\tan\theta.[/tex]
Thus, from the difference of squares identity we have
[tex]\displaystyle{ \frac{(1-\cos\theta)(1+\cos\theta)}{(1-\sin\theta)(1+\sin\theta)}= \frac{1-\cos^2\theta}{1-\sin^2\theta}.[/tex]
From the identity [tex]\sin^2\theta+\cos^2\theta=1[/tex], we have
[tex]\sin^2\theta=1-\cos^2\theta[/tex], and [tex]\cos^2\theta=1-\sin^2\theta[/tex], thus
[tex]\displaystyle{ \frac{1-\cos^2\theta}{1-\sin^2\theta}= \frac{\sin^2\theta}{\cos^2\theta}= (\frac{\sin\theta}{\cos\theta})^2=\tan^2\theta.[/tex]
Answer: [tex]\tan^2\theta.[/tex]
