Question

make the indicated trigonometric substitution in the given algebraic expression and simplify. assume that ( 0 < \theta < pi / 2 ).
( sqrt { x ^ { 2 } - 1 }, quad x = csc ( \theta ) )

Explanation:

Step1: Substitute \(x = \csc(\theta)\)

Substitute \(x=\csc(\theta)\) into \(\sqrt{x^{2}-1}\), we get \(\sqrt{\csc^{2}(\theta)-1}\).

Step2: Use trigonometric identity

Recall the trigonometric identity \(1+\cot^{2}(\theta)=\csc^{2}(\theta)\), so \(\csc^{2}(\theta)-1 = \cot^{2}(\theta)\). Then \(\sqrt{\csc^{2}(\theta)-1}=\sqrt{\cot^{2}(\theta)}\).

Step3: Simplify the square - root

Since \(0 < \theta<\frac{\pi}{2}\), \(\cot(\theta)>0\). So \(\sqrt{\cot^{2}(\theta)}=\cot(\theta)\).

Answer:

\(\cot(\theta)\)