(iv) From the question first we consider Left Hand Side (LHS),

$\sqrt{\left( \left( 1-\cos B \right)/\left( 1+\cos B \right) \right)}$

Then,

$=\sqrt{\left( \left( \left( 1-\cos B \right)/\left( 1+\cos B \right) \right)\left( 1+\cos B \right)/\left( 1+\cos B \right) \right)}$

$=\sqrt{\left( \left( 1-{{\cos }^{2}}B \right)/{{\left( 1+\cos B \right)}^{2}} \right)}$

$=\sqrt{\left( \left( {{\sin }^{2}}B \right)/{{\left( 1+\cos B \right)}^{2}} \right)}$

$=sinB/(1+cosB)$

Then, Right Hand Side (RHS) $=sinB/(1+cosB)$

Therefore, LHS = RHS

(iv)

From the question first we consider Left Hand Side (LHS),

$\sqrt{\left( \left( 1+\cos B \right)/\left( 1-\cos B \right) \right)}$

Then,

$=\sqrt{\left( \left( \left( 1+\cos B \right)/\left( 1-\cos B \right) \right)\left( \left( 1+\cos B \right)/\left( 1+\cos B \right) \right) \right)}$

$=\sqrt{\left( \left( 1+{{\cos }^{2}}B \right)/\left( {{\sin }^{2}}B \right) \right)}$

We know that, $1-{{\cos }^{2}}B={{\sin }^{2}}B$

$=\sqrt{\left( \left( 1+{{\cos }^{2}}B \right)/\left( {{\sin }^{2}}B \right) \right)}$

$=\sqrt{{{\left( \left( 1+\cos B \right)/\left( \sin B \right) \right)}^{2}}}$

$=\sqrt{{{\left( \left( 1/\sin B \right)+\left( \cos B/\sin B \right) \right)}^{2}}}$

$=\sqrt{{{\left( \cos ecB+\cot B \right)}^{2}}}$

$=cosecB+cotB$

Then, Right Hand Side (RHS) $=cosecB+cotB$

Therefore, LHS = RHS