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 $\operatorname{cosec}^{2} A=1+\cot ^{2} A$
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 $\operatorname{cosec}^{2} A=1+\cot ^{2} A$
$\frac{\cos A-\sin A+1}{\cos A+\sin A-1}=\operatorname{cosec} A+\cot A$
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
Show that:
$(i)$ $\tan 48^{\circ} \tan 23^{\circ} \tan 42^{\circ} \tan 67^{\circ}=1$
$(ii)$ $\cos 38^{\circ} \cos 52^{\circ}-\sin 38^{\circ} \sin 52^{\circ}=0$
Show that:
$(i)$ $\tan 48^{\circ} \tan 23^{\circ} \tan 42^{\circ} \tan 67^{\circ}=1$
$(ii)$ $\cos 38^{\circ} \cos 52^{\circ}-\sin 38^{\circ} \sin 52^{\circ}=0$
In $\triangle$ $PQR,$ right-angled at $Q$ (see $Fig.$), $PQ =3 \,cm$ and $PR =6 \,cm$. Determine $\angle QPR$ and $\angle PRQ$.
In $\triangle$ $PQR,$ right-angled at $Q$ (see $Fig.$), $PQ =3 \,cm$ and $PR =6 \,cm$. Determine $\angle QPR$ and $\angle PRQ$.
If $\sec 4 A =\operatorname{cosec}\left( A -20^{\circ}\right),$ where $4 A$ is an acute angle, find the value of $A$. (in $^{\circ}$)
If $\sec 4 A =\operatorname{cosec}\left( A -20^{\circ}\right),$ where $4 A$ is an acute angle, find the value of $A$. (in $^{\circ}$)
$\frac{2 \tan 30^{\circ}}{1-\tan ^{2} 30^{\circ}}=$
$\frac{2 \tan 30^{\circ}}{1-\tan ^{2} 30^{\circ}}=$
$(\sec A+\tan A)(1-\sin A)=..........$
$(\sec A+\tan A)(1-\sin A)=..........$