Answer:
See below
Step-by-step explanation:
[tex]A=\theta[/tex]
[tex]$\sin(90\º - \theta) \cos(90\º - \theta) = \frac{ \tan(\theta)}{1+ \tan^2(\theta)}$[/tex]
According to cofunctions identities:
[tex]\boxed{\sin(90\º - \theta)=\cos(\theta)}[/tex]
[tex]\boxed{\cos(90\º - \theta)=\sin(\theta)}[/tex]
[tex]$\cos(\theta) \sin(\theta) = \frac{ \tan(\theta)}{1+ \tan^2(\theta)}$[/tex]
Once
[tex]$\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)} $[/tex]
[tex]$\cos(\theta) \sin(\theta) = \frac{ \frac{\sin(\theta)}{\cos(\theta)}}{1+ \frac{\sin^2(\theta)}{\cos^2(\theta)}}$[/tex]
Take
[tex]$1+\frac{\sin^2(\theta)}{\cos^2(\theta)}}=\frac{\sin^2(\theta) + \cos^2(\theta)}{\cos^2(\theta)} $[/tex]
[tex]$\cos(\theta) \sin(\theta) = \frac{ \frac{\sin(\theta)}{\cos(\theta)}}{\frac{\sin^2(\theta) + \cos^2(\theta)}{\cos^2(\theta)} }}$[/tex]
[tex]$\cos(\theta) \sin(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \cdot \frac{\cos^2(\theta)}{\sin^2(\theta) + \cos^2(\theta)} }$[/tex]
[tex]$\cos(\theta) \sin(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \cdot \frac{\cos^2(\theta)}{\sin^2(\theta) + \cos^2(\theta)} }$[/tex]
[tex]$\cos(\theta) \sin(\theta) = \frac{\sin(\theta)\cos(\theta)}{\sin^2(\theta) + \cos^2(\theta)} }$[/tex]
Once
[tex]\boxed{\sin^2(\theta) + \cos^2(\theta)=1}[/tex]
[tex]$\cos(\theta) \sin(\theta) = \frac{\sin(\theta)\cos(\theta)}{1 }$[/tex]
[tex]$\cos(\theta) \sin(\theta) = \sin(\theta)\cos(\theta)$[/tex]
[tex]$\therefore \sin(90\º - \theta) \cos(90\º - \theta) = \frac{ \tan(\theta)}{1+ \tan^2(\theta)}$[/tex]
