(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