Explanation

(c) In case of tangent galvanometer as $i = k\tan\phi$
Differentiating both side $w.r.t.\ \phi$
$\frac{di}{d\phi} = k\sec^{2}\phi \Rightarrow di = k\sec^{2}{d\phi}$
$\Rightarrow \frac{di}{i} = \frac{d\phi}{\sin\phi\cos\phi} = \frac{2d\phi}{\sin{2\phi}}$
Hence the error in the measurement will be least when
$\sin{2\phi} = \max{= 1} \Rightarrow 2\phi = 90{^\circ} \Rightarrow \phi = 45{^\circ}$