1−L1+L
L2√31+L
1+L3L2
L(L−1)1−L+1√1−L2
Correct answer is D
Given Cos z = L, z is an acute angle
cot z - cosec zsec z + tan z = cos z
= cos zsin z
cosec z = 1sin z
cot z - cosec z = cos zsin z - 1sin z
cot z - cosec z = L−1sin z
sec z = 1cos z
tan z = sin zcos z
sec z = 1cos z + sin zcos z
= 1l + sin zL
the original eqn. becomes
cot z - cosec zsec z + tan z = L−1sin z1+sinzL
= L(L−1)sin z(1+sin z)
= L(L−1)sin z+1−cos2z
= sin z + 1
= 1 + √1−L2
= L(L−1)1−L+1√1−L2
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