Question

Prove that $\frac{\cos \left(A\right)-\sin \left(A\right)+1}{\cos \left(A\right)+\sin \left(A\right)-1}=\cos ec\left(A\right)+cot\left(A\right)$

Answer 1

Solve for the required proof

Given that $\frac{\cos \left(A\right)-\sin \left(A\right)+1}{\cos \left(A\right)+\sin \left(A\right)-1}=\cos ec\left(A\right)+cot\left(A\right)$

Consider L.H.S

$\frac{\cos \left(A\right)-\sin \left(A\right)+1}{\cos \left(A\right)+\sin \left(A\right)-1}$

Dividing the numerator and denominator by $\sin \left(A\right)$, we get,

$\frac{{\displaystyle \frac{\cos \left(A\right)}{\sin \left(A\right)}}-{\displaystyle \frac{\sin \left(A\right)}{\sin \left(A\right)}}+{\displaystyle \frac{1}{\sin \left(A\right)}}}{{\displaystyle \frac{\cos \left(A\right)}{\sin \left(A\right)}}+{\displaystyle \frac{\sin \left(A\right)}{\sin \left(A\right)}}-{\displaystyle \frac{1}{\sin \left(A\right)}}}$

$=\frac{cot\left(A\right)-1+\cos ec\left(A\right)}{cot\left(A\right)+1-\cos ec\left(A\right)}$ ; $\left[\because cot\left(A\right)=\frac{\cos \left(A\right)}{\sin \left(A\right)},\cos ec\left(A\right)=\frac{1}{\sin \left(A\right)}\right]$

$=\frac{cot\left(A\right)-\left(\cos e{c}^2\left(A\right)-co{t}^2\left(A\right)\right)+\cos ec\left(A\right)}{cot\left(A\right)+1-\cos ec\left(A\right)}$ ;$\left[\because 1+co{t}^2\left(A\right)=\cos e{c}^2\left(A\right)\right]$

$=\frac{cot\left(A\right)+\cos ec\left(A\right)-\left[\left(\cos ec\left(A\right)+cot\left(A\right)\right)\left(\cos ec\left(A\right)-cot\left(A\right)\right)\right]}{cot\left(A\right)+1-\cos ec\left(A\right)}$ ; $\left[\because a^2-b^2=\left(a+b\right)\left(a-b\right)\right]$

$=\frac{\left(cot\left(A\right)+\cos ec\left(A\right)\right)\left[1-\left(\cos ec\left(A\right)-cot\left(A\right)\right)\right]}{cot\left(A\right)+1-\cos ec\left(A\right)}$

$=\frac{\left(cot\left(A\right)+\cos ec\left(A\right)\right)\left[cot\left(A\right)+1-\cos ec\left(A\right)\right]}{cot\left(A\right)+1-\cos ec\left(A\right)}$

$=cot\left(A\right)+\cos ec\left(A\right)$

$=$R.H.S

$\Rightarrow$L.H.S$=$R.H.S

Hence, the given identity is proved.