Question

SinA /1+CosA + 1+(1+CosA)^2/ SinA = 2CosecA

Answer 1

Answer:

$\displaystyle{\dfrac{\sin A}{1+\cos A} +1+ \dfrac{(1+\cos A)^2}{\sin A}=2\csc+1 \ \text{Proved}}$

Step-by-step explanation:

Given :

$\displaystyle{\dfrac{\sin A}{1+\cos A} +1+ \frac{(1+\cos A)^2}{\sin A}=2\csc+1 }$

$\displaystyle{\text{L.H..S}=\dfrac{\sin A}{1+\cos A} +1+ \frac{(1+\cos A)^2}{\sin A} }\\\\\\\displaystyle{\rightarrow \dfrac{\sin A}{1+\cos A} +1+ \frac{(1+\cos A)^2}{\sin A} }\\\\\\\displaystyle{\rightarrow \dfrac{\sin^2 A+\sin A(1+\cos A)+(1+\cos A)^2}{\sin A(1+\cos A)}} \\\\\\\displaystyle{\rightarrow \dfrac{\sin^2 A+\sin A(1+\cos A)+(1+\cos^2 A+2\cos A)}{\sin A(1+\cos A)}} \\\\\\\displaystyle{\rightarrow \dfrac{1+1+2\cos A+\sin A(1+\cos A)}{\sin A(1+\cos A)}}$

$\leq \displaystyle{\rightarrow \dfrac{2+2\cos A+\sin A(1+\cos A)}{\sin A(1+\cos A)}}\\\\\\\displaystyle{\rightarrow \dfrac{2(1+\cos A)+\sin A(1+\cos A)}{\sin A(1+\cos A)}} \\\\\\\displaystyle{\rightarrow \dfrac{2+\sin A}{\sin A}}\\\\\\\displaystyle{\rightarrow \dfrac{2}{\sin A}+1}\\\\\\\displaystyle{\rightarrow 2 \csc+1}\\\\\\\display \large \text{L.H.S. = R.H.S. , Hence proved}$