(v)
From the question firstly we consider Left Hand Side (LHS),
$(1+cotA–cosecA)(1+tanA+secA)$
We know that,
$cotA=sinA/cosA$, $cosecA=1/cosA$, $tanA=cosA/sinA$, $secA=1/sinA$
$(1+(sinA/cosA)+(1/cosA))(1+(cosA/sinA)–(1/sinA))$
Taking LCM we get,
$((cosA+sinA+1)/cosA)((sinA+cosA–1)/sinA)$
$\left( {{\left( \sin A+\cos A \right)}^{2}}-{{1}^{2}} \right)/\left( \sin A\cos A \right)$
$=(1+2sinAcosA–1)/(sinAcosA)$
$=(2sinAcosA)/(sinAcosA)$
By simplification we get,
$=2$
Then, Right Hand Side (RHS) $=2$
Therefore, LHS = RHS
(vi)
From the question firstly we consider Left Hand Side (LHS),
= $sinAcotA+sinAcosecA$
We know that, $cotA=cosA/sinA$, $cosecA=1/sinA$
= $sinA(cosA/sinA)+sinA(1/sinA)$
= $cosA+1$
Then, Right Hand Side (RHS) $=1+cosA$
Therefore, LHS = RHS