Given that cos z = L , whrere z is an acute angle, find an expression for \(\frac{\cot z - \csc z}{\sec z + \tan z}\)
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Correct Answer: Option D
Explanation:
Given Cos z = L, z is an acute angle
\(\frac{\text{cot z - cosec z}}{\text{sec z + tan z}}\) = cos z
= \(\frac{\text{cos z}}{\text{sin z}}\)
cosec z = \(\frac{1}{\text{sin z}}\)
cot z - cosec z = \(\frac{\text{cos z}}{\text{sin z}}\) - \(\frac{1}{\text{sin z}}\)
cot z - cosec z = \(\frac{L - 1}{\text{sin z}}\)
sec z = \(\frac{1}{\text{cos z}}\)
tan z = \(\frac{\text{sin z}}{\text{cos z}}\)
sec z = \(\frac{1}{\text{cos z}}\) + \(\frac{\text{sin z}}{\text{cos z}}\)
= \(\frac{1}{l}\) + \(\frac{\text{sin z}}{L}\)
the original eqn. becomes
\(\frac{\text{cot z - cosec z}}{\text{sec z + tan z}}\) = \(\frac{L - \frac{1}{\text{sin z}}}{1 + sin \frac{z}{L}}\)
= \(\frac{L(L - 1)}{\text{sin z}(1 + \text{sin z})}\)
= \(\frac{L(L - 1)}{\text{sin z} + 1 - cos^2 z}\)
= sin z + 1
= 1 + \(\sqrt{1 - L^2}\)
= \(\frac{L(L - 1)}{1 - L + 1 \sqrt{1 - L^2}}\)
Given Cos z = L, z is an acute angle
\(\frac{\text{cot z - cosec z}}{\text{sec z + tan z}}\) = cos z
= \(\frac{\text{cos z}}{\text{sin z}}\)
cosec z = \(\frac{1}{\text{sin z}}\)
cot z - cosec z = \(\frac{\text{cos z}}{\text{sin z}}\) - \(\frac{1}{\text{sin z}}\)
cot z - cosec z = \(\frac{L - 1}{\text{sin z}}\)
sec z = \(\frac{1}{\text{cos z}}\)
tan z = \(\frac{\text{sin z}}{\text{cos z}}\)
sec z = \(\frac{1}{\text{cos z}}\) + \(\frac{\text{sin z}}{\text{cos z}}\)
= \(\frac{1}{l}\) + \(\frac{\text{sin z}}{L}\)
the original eqn. becomes
\(\frac{\text{cot z - cosec z}}{\text{sec z + tan z}}\) = \(\frac{L - \frac{1}{\text{sin z}}}{1 + sin \frac{z}{L}}\)
= \(\frac{L(L - 1)}{\text{sin z}(1 + \text{sin z})}\)
= \(\frac{L(L - 1)}{\text{sin z} + 1 - cos^2 z}\)
= sin z + 1
= 1 + \(\sqrt{1 - L^2}\)
= \(\frac{L(L - 1)}{1 - L + 1 \sqrt{1 - L^2}}\)