(i)
From the question first we consider Left Hand Side (LHS),
$=\sqrt{\left( \cos e{{c}^{2}}a-1 \right)}$
We know that, $\cos e{{c}^{2}}a-1={{\cot }^{2}}a$
$=\sqrt{\left( {{\cot }^{2}}a \right)}$
Then,
$=cota$
$=cosa/sina$
$=cosa(1/sina)$
Also we know that, $1/sina=coseca$
$=cosacoseca$
Then, Right Hand Side (RHS) $=cosacoseca$
Therefore, LHS = RHS
(ii)
From the question first we consider Left Hand Side (LHS),
$\sqrt{\left( \left( 1+\sin a \right)/\left( 1-\sin a \right) \right)}+\sqrt{\left( \left( 1-\sin a \right)/\left( 1+\sin a \right) \right)}$
Then,$=\sqrt{\left( \left( \left( 1+\sin a \right)/\left( 1-\sin a \right) \right)\left( \left( 1+\sin a \right)/\left( 1+\sin a \right) \right) \right)}+\sqrt{\left( \left( \left( 1-\sin a \right)/\left( 1+\sin a \right) \right)\left( \left( 1+\sin a \right)\left( 1+\sin a \right) \right) \right)}$
$=\sqrt{\left( {{\left( 1+\sin a \right)}^{2}}/\left( 1-{{\sin }^{2}}a \right) \right)}+\sqrt{\left( {{\left( 1-\sin a \right)}^{2}}/\left( 1-{{\sin }^{2}}a \right) \right)}$
We know that, $1-{{\sin }^{2}}a={{\cos }^{2}}a$
$=\sqrt{\left( {{\left( 1+\sin a \right)}^{2}}/{{\cos }^{2}}a \right)}+\sqrt{\left( {{\left( 1-\sin a \right)}^{2}}/{{\cos }^{2}}a \right)}$
$=((1+sina)/cosa)+((1–sina)/cosa)$
$=2/cosa$
$=2seca$
Then, Right Hand Side (RHS) $=2seca$
Therefore, LHS = RHS