Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

$\frac{\cos A-\sin A+1}{\cos A+\sin A-1}=\operatorname{cosec} A+\cot A,$ using the identity $\operatorname{cosec}^{2} A=1+\cot ^{2} A$

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

Using the identity $\operatorname{cosec}^{2} A =1+\cot ^{2} A$

$L.H.S.$ $=\frac{\cos A -\sin A +1}{\cos A +\sin A -1}$

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

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

$=\frac{\{(\cot A)-(1-\operatorname{cosec} A)\}\{(\cot A)-(1-\operatorname{cosec} A)\}}{\{(\cot A)+(1-\operatorname{cosec} A)\}\{(\cot A)-(1-\operatorname{cosec} A)\}}$

$=\frac{(\cot A-1+\operatorname{cosec} A)^{2}}{(\cot A)^{2}-(1-\operatorname{cosec} A)^{2}}$

$=\frac{\cot ^{2} A+1+\operatorname{cosec}^{2} A-2 \cot A-2 \operatorname{cosec} A+2 \cot A \operatorname{cosec} A}{\cot ^{2} A-\left(1+\operatorname{cosec}^{2} A-2 \operatorname{cosec} A\right)}$

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

$=\frac{2 \operatorname{cosec} A(\operatorname{cosec} A+\cot A)-2(\cot A+\operatorname{cosec} A)}{\cot ^{2} A-\operatorname{cosec}^{2} A-1+2 \operatorname{cosec} A}$

$=\frac{(\operatorname{cosec} A+\cot A)(2 \operatorname{cosec} A-2)}{-1-1+2 \operatorname{cosec} A}$

$=\frac{(\operatorname{cosec} A+\cot A)(2 \operatorname{cosec} A-2)}{(2 \operatorname{cosec} A-2)}$

$=\operatorname{cosec} A+\cot A$

$= R . H.S.$

Similar Questions

In $\triangle$ $OPQ$, right-angled at $P$, $OP =7\, cm$ and $OQ - PQ =1\, cm$ (see $Fig.$). Determine the values of $\sin Q$ and $\cos Q$.

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$\frac{\sin 30^{\circ}+\tan 45^{\circ}-\operatorname{cosec} 60^{\circ}}{\sec 30^{\circ}+\cos 60^{\circ}+\cot 45^{\circ}}$

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