While proving $\dfrac{\sin heta-\cos heta+1}{\sin heta+\cos heta-1…

Question

While proving $\dfrac{\sin	heta-\cos	heta+1}{\sin	heta+\cos	heta-1}=\dfrac{1}{\sec	heta-	an	heta}$, why do we divide the numerator and denominator of the left-hand side by $\cos	heta$?

Accepted Answer

A. To convert the expression into terms of $\sec	heta$ and $	an	heta$

The target expression on the right involves $\sec	heta$ and $	an	heta$. Dividing by $\cos	heta$ converts $\frac{\sin	heta}{\cos	heta}$ into $	an	heta$, $\frac{1}{\cos	heta}$ into $\sec	heta$, and $\frac{\cos	heta}{\cos	heta}$ into 1, so the LHS can be compared directly with the RHS.

Answer

To make both numerator and denominator equal to 1

Answer

To convert the expression into terms of $\cosec	heta$ and $\cot	heta$

Answer

Because $\cos	heta=1$ for every acute angle

While proving $\dfrac{\sin\theta-\cos\theta+1}{\sin\theta+\cos\theta-1}=\dfrac{1}{\sec\theta-\tan\theta}$, why do we divide the numerator and denominator of the left-hand side by $\cos\theta$?