Given that cos z = L , whrere z is an acute…

Question 1 Report

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

1-L

1+L

L²√3

1+L

1+L³

L²

L(L−1)

1−L+1√1−L²

Answer Details

Given Cos z = L, z is an acute angle
$\frac{cot z - cosec z}{sec z + tan z}$ = cos z
= $\frac{cos z}{sin z}$
cosec z = $\frac{1}{sin z}$
cot z - cosec z = $\frac{cos z}{sin z}$ - $\frac{1}{sin z}$
cot z - cosec z = $\frac{L - 1}{sin z}$
sec z = $\frac{1}{cos z}$
tan z = $\frac{sin z}{cos z}$
sec z = $\frac{1}{cos z}$ + $\frac{sin z}{cos z}$
= $\frac{1}{l}$ + $\frac{sin z}{L}$
the original eqn. becomes
$\frac{cot z - cosec z}{sec z + tan z}$ = $\frac{L - \frac{1}{sin z}}{1 + s i n \frac{z}{L}}$
= $\frac{L (L - 1)}{sin z (1 + sin z)}$
= $\frac{L (L - 1)}{sin z + 1 - c o s^{2} z}$
= sin z + 1
= 1 + $\sqrt{1 - L^{2}}$
= $\frac{L (L - 1)}{1 - L + 1 \sqrt{1 - L^{2}}}$

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

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

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