Prove the following identities:$ \frac{\cos A \operatorname{cosec} A-\sin A \sec A}{\cos A+\sin A}=\operatorname{cosec} A-\sec A $

To do:

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

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,

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

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

$=\frac{\cos ^{2} A-\sin ^{2} A}{\sin A \cos A(\cos A+\sin A)}$

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

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

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

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

$=\operatorname{cosec} A-\sec A$

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Updated on: 10-Oct-2022

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