Question
Prove the following trigonometric identities.
$\frac{\cos\text{A cosec A}-\sin\text{A}\sec\text{A}}{\cos\text{A}+\sin\text{A}}=\text{cosec A}-\sec\text{A}$

Answer

We have to prove $\frac{\cos\text{A cosec A}-\sin\text{A}\sec\text{A}}{\cos\text{A}+\sin\text{A}}=\text{cosec A}-\sec\text{A}$
So,
$\text{L.H.S}=\frac{\cos\text{A cosec A}-\sin\text{A}\sec\text{A}}{\cos\text{A}+\sin\text{A}}=\frac{\cos\text{A}\frac{1}{\sin\text{A}}-\sin\text{A}\frac{1}{\cos\text{A}}}{\cos\text{A}+\sin\text{A}}$
$=\frac{\frac{\cos^2\text{A}-\sin^2\text{A}}{\sin\text{A}\cos\text{A}}}{\cos\text{A}+\sin\text{A}}$
$=\frac{\cos^2\text{A}-\sin^2\text{A}}{\sin\text{A}\cos\text{A}(\cos\text{A}+\sin\text{A})}$
$=\frac{(\cos\text{A}-\sin\text{A})(\cos\text{A}+\sin\text{A})}{\sin\text{A}\cos\text{A}(\cos\text{A}+\sin\text{A})}$
$=\frac{\cos\text{A}-\sin\text{A}}{\sin\text{A}\cos\text{A}}$
$=\frac{\cos\text{A}}{\sin\text{A}\cos\text{A}}-\frac{\sin\text{A}}{\sin\text{A}\cos\text{A}}$
$=\frac{1}{\sin\text{A}}-\frac{1}{\cos\text{A}}$
$=\text{cosec A}-\sec\text{A}=\text{R.H.S}$
$\therefore\ \text{L.H.S}=\text{R.H.S}$

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