Question

Prove that ln(cscu−cotu)=−ln(cscu+cotu)

Answer 1

Given : ln(cscu−cotu)=−ln(cscu+cotu)

To Find : Prove

Solution:

ln(cscu−cotu)=−ln(cscu+cotu)

LHS = ln(cscu−cotu)

= ln(1/Sinu − cosu/sinu)

= ln ( 1 - cosu)/(sin u))

multiply numerator and denominator by 1 + cosu   inside ln

= ln ( 1 - cosu)(1 + cosu)/(1 + Cosu)(sin u))

=  ln ( 1 - cos²u)/(1 + Cosu)(sin u))

=  ln ( sin²u)/(1 + Cosu)(sin u))

= ln ( sinu)/(1 + Cosu))

= ln ( 1/(1/Sinu + Cosu/Sinu)

= ln( 1/(cscu + cotu))

= ln ((cscu + cot u)⁻¹)

= - ln(cscu+cotu)

= RHS

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Answer 2

SOLUTION

TO PROVE

$\sf{ \ln ( \csc u \: - \cot u) = - \ln ( \csc u \: + \cot u) }$

FORMULA TO BE IMPLEMENTED

$\sf{1. \: { \csc}^{2}u - { \cot}^{2} u = 1 }$

$\sf{2. \: \ln (xy) = \ln x + \ln y}$

EVALUATION

Here we have

$\sf{ { \csc}^{2}u - { \cot}^{2} u = 1 }$

Taking logarithm in both sides we get

$\sf{ \implies \ln ( { \csc}^{2}u - { \cot}^{2} u) = \ln 1 }$

$\sf{ \implies \ln \bigg[( \csc u + \cot u)( \csc u - \cot u) \bigg] = 0 }$

$\sf{ \implies \ln ( \csc u + \cot u) + \ln ( \csc u - \cot u) = 0 }$

$\sf{ \implies \ln ( \csc u + \cot u) = - \ln ( \csc u - \cot u) }$

∴ $\sf{ \ln ( \csc u \: - \cot u) = - \ln ( \csc u \: + \cot u) }$

Hence proved

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