(d) Given, $\tan \theta = \frac{1}{7},\sin \phi = \frac{1}{{\sqrt {10} }}$

$\sin \theta = \frac{1}{{\sqrt {50} }},\,\,\cos \theta = \frac{7}{{\sqrt {50} }},\,\,\cos \phi = \frac{3}{{\sqrt {10} }}$

$\therefore \,\,\cos 2\phi  = 2{\cos ^2}\phi – 1 = 2.\frac{9}{{10}} – 1 = \frac{8}{{10}}$

$\sin 2\phi = 2\sin \phi \cos \phi = 2 \times .\frac{1}{{\sqrt {10} }} \times \frac{3}{{\sqrt {10} }} = \frac{6}{{10}}$

$\therefore \cos (\theta + 2\phi ) = \cos \theta \cos 2\phi – \sin \theta \sin 2\phi $

$ = \frac{7}{{\sqrt {50} }} \times \frac{8}{{10}} – \frac{1}{{\sqrt {50} }}.\frac{6}{{10}}$

$ = \frac{{56 – 6}}{{10\sqrt {50} }} = \frac{{50}}{{10\sqrt {50} }} = \frac{{5\sqrt 2 }}{{10}} = \frac{1}{{\sqrt 2 }}$

$\therefore \theta + 2\phi = {45^o}$.