Question

$\frac{\sin A}{\sin ({90}^o-A)}+\frac{\cos A}{\cos ({90}^o-A)}=\sec ({90}^o-A)co\sec ({90}^o-A)$

Answer 1

We substitute trignometric identity $sin({90}^o-A)=cosA$ and $cos({90}^o-A)=sinA$.

Substituting these identities in the above equation,

$\frac{sinA}{cosA}+\frac{cosA}{sinA}$

$=\frac{si{n}^{2}A+co{s}^{2}A}{cosAsinA}$

$=\frac{1}{cosAsinA}$

$=secAcosecA$ which can be rewritten using identities $secA=cosec(90-A)$ and $cosecA=sec(90-A)$

Hence the given identity is written as $sec(90-A)cosec(90-A)$