(vii)
From the question first we consider Left Hand Side (LHS),
$=cosecB+cotB$
Now multiply and divide by $cosecB–cotB$ we get,
$=((cosecB+cotB)/1)((cosecB–cotB)/(cosecB–cotB))$
By cross multiplication we get,
$=\left( \cos e{{c}^{2}}B-{{\cot }^{2}}B \right)/\left( \cos ecB-\cot B \right)$
$=\left( 1+{{\cot }^{2}}B-{{\cot }^{2}}B \right)/\left( \cos ecB-\cot B \right)$
$=(1/(cosecB–cotB))$
Then, Right Hand Side (RHS) $=(1/(cosecB–cotB))$
Therefore, LHS = RHS
(viii)
From the question first we consider Left Hand Side (LHS),
$=((cosecB)/(cosecB–1))+((cosecB)/(cosecB+1))$
By cross multiplication we get,
$\left( \cos e{{c}^{2}}B+\cos ecB+\cos e{{c}^{2}}B-\cos ecB \right)/\left( \cos e{{c}^{2}}B-1 \right)$
We know that, $\cos e{{c}^{2}}B-1={{\cot }^{2}}B$
$=2\cos e{{c}^{2}}B/{{\cot }^{2}}B$
Also we know that, $\cos e{{c}^{2}}B=1/{{\sin }^{2}}B$
$=\left( 2/{{\sin }^{2}}B \right)/\left( {{\cos }^{2}}B/{{\sin }^{2}}B \right)$
$=2/{{\cos }^{2}}B$
$=2{{\sec }^{2}}B$
Then, Right Hand Side (RHS) $=2{{\sec }^{2}}B$
Therefore, LHS = RHS