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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.