Prove that:$ \frac{\cot A-\cos A}{\cot A+\cos A}=\frac{\operatorname{cosec} A-1}{\operatorname{cosec} A+1} $

To do:

We have to prove that \( \frac{\cot A-\cos A}{\cot A+\cos A}=\frac{\operatorname{cosec} A-1}{\operatorname{cosec} A+1} \).

Solution:

We know that,

$\sin^2 A+\cos^2 A=1$

$\operatorname{cosec}^2 A-\cot^2 A=1$

$\sec^2 A-\tan^2 A=1$

$\cot A=\frac{\cos A}{\sin A}$

$\tan A=\frac{\sin A}{\cos A}$

$\operatorname{cosec} A=\frac{1}{\sin A}$

$\sec A=\frac{1}{\cos A}$

Therefore,

Let us consider LHS,

$\frac{\cot \mathrm{A}-\cos \mathrm{A}}{\cot \mathrm{A}+\cos \mathrm{A}}=\frac{\frac{\cos \mathrm{A}}{\sin \mathrm{A}}-\cos \mathrm{A}}{\frac{\cos \mathrm{A}}{\sin \mathrm{A}}+\cos \mathrm{A}}$

$=\frac{\frac{\cos \mathrm{A}-\sin \mathrm{A} \cos \mathrm{A}}{\sin \mathrm{A}}}{\cos \mathrm{A}+\sin \mathrm{A} \cos \mathrm{A}}{\sin \mathrm{A}}$

$=\frac{\cos \mathrm{A}(1-\sin \mathrm{A})}{\cos \mathrm{A}(1+\sin \mathrm{A})}$

$=\frac{1-\sin \mathrm{A}}{1+\sin \mathrm{A}}$

Let us consider RHS,

$\frac{\operatorname{cosec} A-1}{\operatorname{cosec} A+1}=\frac{\frac{1}{\sin A}-1}{\frac{1}{\sin A}+1}$

$=\frac{\frac{1-\sin A}{\sin A}}{\frac{1+\sin A}{\sin A}}$

$=\frac{1-\sin A}{1+\sin A}$

Here,

LHS $=$ RHS

Hence proved.    

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Updated on: 10-Oct-2022

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