Prove that

frac{sin theta-cos theta+1}{sin | vedclass

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Question

Since we will apply the identity involving $\sec 	heta$ and $	an 	heta,$ let us first convert the $LHS$ (of the identity we need to prove) in terms

Vedclass · Accepted Answer

Vedclass · Answer

Since we will apply the identity involving $\sec 	heta$ and $	an 	heta,$ let us first convert the $LHS$ (of the identity we need to prove) in terms of $\sec 	heta$ and $	an 	heta$ by dividing numerator and denominator by $\cos 	heta .$

$LHS=\frac{\sin 	heta-\cos 	heta+1}{\sin 	heta+\cos 	heta-1}=\frac{	an 	heta-1+\sec 	heta}{	an 	heta+1-\sec 	heta}$

$=\frac{(	an 	heta+\sec 	heta)-1}{(	an 	heta-\sec 	heta)+1}=\frac{\{(	an 	heta+\sec 	heta)-1\}(	an 	heta-\sec 	heta)}{\{(	an 	heta-\sec 	heta)+1\}(	an 	heta-\sec 	heta)}$

$=\frac{\left(	an ^{2} 	heta-\sec ^{2} 	hetaight)-(	an 	heta-\sec 	heta)}{\{	an 	heta-\sec 	heta+1\}(	an 	heta-\sec 	heta)}$

$=\frac{-1-	an 	heta+\sec 	heta}{(	an 	heta-\sec 	heta+1)(	an 	heta-\sec 	heta)}$

$=\frac{-1}{	an 	heta-\sec 	heta}=\frac{1}{\sec 	heta-	an 	heta}$

which is the RHS of the identity, we are required to prove.

Prove that

$\frac{\sin \theta-\cos \theta+1}{\sin \theta+\cos \theta-1}=\frac{1}{\sec \theta-\tan \theta},$ using the identity

$\sec ^{2} \theta=1+\tan ^{2} \theta$

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