Prove that:
\[\begin{align}
  & (i)\sqrt{\dfrac{\sec \theta -1}{\sec \theta +1}}+\sqrt{\dfrac{\sec \theta +1}{\sec \theta -1}}=2cosec\theta \\
 & (ii)\sqrt{\dfrac{1+\sin \theta }{1-\sin \theta }}+\sqrt{\dfrac{1-\sin \theta }{1+\sin \theta }}=2\sec \theta \\
\end{align}\]

Hint: To solve the given question, we should know some of the trigonometric properties that are given below, we should know that sine and cosine are inverse of secant and cosecant function respectively, that is \[\sec \theta =\dfrac{1}{\cos \theta }\And \csc \theta =\dfrac{1}{\sin \theta }\]. Also, we should know the trigonometric identity relation between sine-cosine function, and secant-tangent functions,
\[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\And 1+{{\tan }^{2}}\theta ={{\sec }^{2}}\theta \].

Complete step by step answer:
The first statement we need to prove is, \[\sqrt{\dfrac{\sec \theta -1}{\sec \theta +1}}+\sqrt{\dfrac{\sec \theta +1}{\sec \theta -1}}=2cosec\theta \]. The LHS of the statement is \[\sqrt{\dfrac{\sec \theta -1}{\sec \theta +1}}+\sqrt{\dfrac{\sec \theta +1}{\sec \theta -1}}\], and the RHS is \[2cosec\theta \]. Multiplying the first term in the LHS by \[\sqrt{\dfrac{\sec \theta -1}{\sec \theta -1}}\], and the second term by \[\sqrt{\dfrac{\sec \theta +1}{\sec \theta +1}}\]. We get
\[\Rightarrow \sqrt{\dfrac{\sec \theta -1}{\sec \theta +1}}\sqrt{\dfrac{\sec \theta -1}{\sec \theta -1}}+\sqrt{\dfrac{\sec \theta +1}{\sec \theta -1}}\sqrt{\dfrac{\sec \theta +1}{\sec \theta +1}}\]
Simplifying the above expression, it can be written as
\[\Rightarrow \sqrt{\dfrac{{{\left( \sec \theta -1 \right)}^{2}}}{{{\left( \sec \theta \right)}^{2}}-{{\left( 1 \right)}^{2}}}}+\sqrt{\dfrac{{{\left( \sec \theta +1 \right)}^{2}}}{{{\left( \sec \theta \right)}^{2}}-{{\left( 1 \right)}^{2}}}}\]
Using the trigonometric identity, \[1+{{\tan }^{2}}\theta ={{\sec }^{2}}\theta \], the denominator of the above expression can be expressed as
\[\Rightarrow \sqrt{\dfrac{{{\left( \sec \theta -1 \right)}^{2}}}{{{\tan }^{2}}\theta }}+\sqrt{\dfrac{{{\left( \sec \theta +1 \right)}^{2}}}{{{\tan }^{2}}\theta }}\]
canceling out the power, we get
\[\Rightarrow \dfrac{\left( \sec \theta -1 \right)}{\tan \theta }+\dfrac{\left( \sec \theta +1 \right)}{\tan \theta }\]
As the denominator of both terms is the same, we can add the numerator directly, by doing this we get
\[\Rightarrow \dfrac{2\sec \theta }{\tan \theta }\]
We know that \[\sec \theta =\dfrac{1}{\cos \theta }\And \tan \theta =\dfrac{\sin \theta }{\cos \theta }\]. Using this, the above expression can be simplified as
\[\Rightarrow \dfrac{2}{\sin \theta }=2\csc \theta =RHS\]
\[\therefore LHS=RHS\]
Hence, proved.

The second statement to prove is \[\sqrt{\dfrac{1+\sin \theta }{1-\sin \theta }}+\sqrt{\dfrac{1-\sin \theta }{1+\sin \theta }}=2\sec \theta \]. The LHS of the statement is \[\sqrt{\dfrac{1+\sin \theta }{1-\sin \theta }}+\sqrt{\dfrac{1-\sin \theta }{1+\sin \theta }}\], and the RHS of the statement is \[2\sec \theta \].
Multiplying the first term in the LHS by \[\sqrt{\dfrac{1+\sin \theta }{1+\sin \theta }}\], and second term by \[\sqrt{\dfrac{1-\sin \theta }{1-\sin \theta }}\]. We get
\[\sqrt{\dfrac{1+\sin \theta }{1-\sin \theta }}\sqrt{\dfrac{1+\sin \theta }{1+\sin \theta }}+\sqrt{\dfrac{1-\sin \theta }{1+\sin \theta }}\sqrt{\dfrac{1-\sin \theta }{1-\sin \theta }}\]
Simplifying the above expression, we get
 \[\Rightarrow \sqrt{\dfrac{{{\left( 1+\sin \theta \right)}^{2}}}{{{1}^{2}}-{{\sin }^{2}}\theta }}+\sqrt{\dfrac{{{\left( 1-\sin \theta \right)}^{2}}}{{{1}^{2}}-{{\sin }^{2}}\theta }}\]
Using the trigonometric identity, \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\], the denominator of the above expression can be expressed as
\[\Rightarrow \sqrt{\dfrac{{{\left( 1+\sin \theta \right)}^{2}}}{{{\cos }^{2}}\theta }}+\sqrt{\dfrac{{{\left( 1-\sin \theta \right)}^{2}}}{{{\cos }^{2}}\theta }}\]
canceling out the square and square root from the above expression, we get
\[\Rightarrow \dfrac{\left( 1+\sin \theta \right)}{\cos \theta }+\dfrac{\left( 1-\sin \theta \right)}{\cos \theta }\]
 As the denominator of both terms is the same, we can add the numerator directly, by doing this we get
\[\Rightarrow \dfrac{2}{\cos \theta }\]
Using \[\sec \theta =\dfrac{1}{\cos \theta }\], the above expression can be written as
\[\Rightarrow 2\sec \theta =RHS\]
\[\therefore LHS=RHS\]
Hence, proved.

Note: To solve problems based on trigonometric functions, one should remember the trigonometric properties and the identities. The properties we used to prove the given statements are \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\And 1+{{\tan }^{2}}\theta ={{\sec }^{2}}\theta \], also \[\sec \theta =\dfrac{1}{\cos \theta }\And \csc \theta =\dfrac{1}{\sin \theta }\].