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  • Prove that :  <img alt="\frac{\sin \theta - \cos \theta +1}{\cos \theta + \sin \theta -1 }= \frac{1}{\sec \theta - \tan \theta }" src="https://learn.careers360.com/latex-image/?%5Cdpi%7B100%7D%20%5Cfrac%7B%5Csin%20%5Ctheta%20-%20%5Ccos%2

Prove that : 

\frac{\sin \theta - \cos \theta +1}{\cos \theta + \sin \theta -1 }= \frac{1}{\sec \theta - \tan \theta }

 

 
 
 
 
 

Answers (1)

 

\\\frac{\sin \theta-\cos \theta+1}{\cos \theta+\sin \theta-1}=\frac{1}{\sec \theta-\tan \theta} \\\\ \text { L.H.S }=\frac{\sin \theta-\cos \theta+1}{\cos \theta+\sin \theta-1} \\ \\ \text { Dividing both numerator and denominator by } \cos \theta $ we get $\\ =\frac{\tan \theta-1+\sec \theta}{1+\tan \theta-\sec \theta}\\\\ =\frac{\tan \theta+\sec \theta-1}{\left(\sec ^{2} \theta-\tan ^{2} \theta\right)+\tan \theta-\sec \theta} \\ =\frac{\tan \theta+\sec \theta-1}{(\sec \theta-\tan \theta)(\sec \theta+\tan \theta-1)} \\ =\frac{1}{\sec \theta-\tan \theta}=\operatorname{R.H} S

Hence proved

Posted by

Safeer PP

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