Question

(iii) sin (90° - A) cos (90° - A)=tanA/
1+tan²A​

Answer 1

See below

Explanation:

`A=\theta`

`$\sin(90\º - \theta) \cos(90\º - \theta) = ( \tan(\theta))/(1+ \tan^2(\theta))$`

According to cofunctions identities:

`\boxed{\sin(90\º - \theta)=\cos(\theta)}`

`\boxed{\cos(90\º - \theta)=\sin(\theta)}`

`$\cos(\theta) \sin(\theta) = ( \tan(\theta))/(1+ \tan^2(\theta))$`

Once

`$\tan(\theta)=(\sin(\theta))/(\cos(\theta)) $`

`$\cos(\theta) \sin(\theta) = ( (\sin(\theta))/(\cos(\theta)))/(1+ (\sin^2(\theta))/(\cos^2(\theta)))$`

Take

`$1+(\sin^2(\theta))/(\cos^2(\theta))}=(\sin^2(\theta) + \cos^2(\theta))/(\cos^2(\theta)) $`

`$\cos(\theta) \sin(\theta) = ( (\sin(\theta))/(\cos(\theta)))/((\sin^2(\theta) + \cos^2(\theta))/(\cos^2(\theta)) )}$`

`$\cos(\theta) \sin(\theta) = (\sin(\theta))/(\cos(\theta)) \cdot (\cos^2(\theta))/(\sin^2(\theta) + \cos^2(\theta)) }$`

`$\cos(\theta) \sin(\theta) = (\sin(\theta))/(\cos(\theta)) \cdot (\cos^2(\theta))/(\sin^2(\theta) + \cos^2(\theta)) }$`

`$\cos(\theta) \sin(\theta) = (\sin(\theta)\cos(\theta))/(\sin^2(\theta) + \cos^2(\theta)) }$`

Once

`\boxed{\sin^2(\theta) + \cos^2(\theta)=1}`

`$\cos(\theta) \sin(\theta) = (\sin(\theta)\cos(\theta))/(1 )$`

`$\cos(\theta) \sin(\theta) = \sin(\theta)\cos(\theta)$`

`$\therefore \sin(90\º - \theta) \cos(90\º - \theta) = ( \tan(\theta))/(1+ \tan^2(\theta))$`