(v)

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

$=tanB–cotB$

We know that, $tanB=sinB/cosB$, $cotB=cosB/sinB$

Then,

$=(sinB/cosB)–(cosB/sinB)$

$=\left( {{\sin }^{2}}B-{{\cos }^{2}}B \right)/\left( \sin B\cos B \right)$

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

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

So,

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

Then, Right Hand Side (RHS) $=\left( 1-2{{\cos }^{2}}B \right)/\left( \sin B\cos B \right)$

Therefore, LHS = RHS

(vi)

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

$=\left( \left( 1+{{\tan }^{2}}B \right)\cot B \right)/\cos e{{c}^{2}}B$

We know that, $1+{{\tan }^{2}}B={{\sec }^{2}}$

$=\left( {{\sec }^{2}}\cot B \right)/\cos e{{c}^{2}}B$

Also we know that, ${{\sec }^{2}}B=1/{{\cos }^{2}}B,\cot B=\cos B/\sin B$

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

$=\left( 1/\left( \cos A\sin A \right) \right)/\left( 1/{{\sin }^{2}}A \right)$

$=sinB/cosB$

$=tanB$

Then, Right Hand Side (RHS) $=tanB$

Therefore, LHS = RHS