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Given that cos z = L , whrere z is an acute angle, find an e...

Given that cos z = L , whrere z is an acute angle, find an expression for \(\frac{\cot z - \csc z}{\sec z + \tan z}\)

A.

\(\frac{1 - L}{1 + L}\)

B.

\(\frac{L^2 \sqrt{3}}{1 + L}\)

C.

\(\frac{1 + L^3}{L^2}\)

D.

\(\frac{L(L - 1)}{1 - L + 1 \sqrt{1 - L^2}}\)

Correct answer is D

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 + \(\sqrt{1 - L^2}\)

= \(\frac{L(L - 1)}{1 - L + 1 \sqrt{1 - L^2}}\)