Question

prove that sin theta minus cos theta + 1 / sin theta + cos theta minus 1 is equals to 1 / secant theta minus tan theta
(using the identities secant squared theta is equals to 1 + tan squared theta )​

Answer 1

Refer the attachment!

Answer 2

$\huge{\sf{\blue{Answer}}}$

$\huge{\sf{\blue{\displaystyle{\frac{(\sin \theta - \cos \theta +1 )}{(\sin \theta +\cos \theta -1)} = \left(\frac{1}{\sec \theta - \tan \theta}\right) \ Proved}}}}$

Step-by-step explanation:

Given ;

$\displaystyle{\frac{(\sin \theta - \cos \theta +1 )}{(\sin \theta +\cos \theta -1)} = \frac{1}{\sec \theta - \tan \theta}}\\\\\\ \displaystyle{L.H.S.=\left(\frac{\sin \theta - \cos \theta +1 }{\sin \theta +\cos \theta -1}\right)}\\\\\\ \displaystyle{Dividing \ by \ \cos\theta \ in \ L.H.S \ we \ get}\\\\\\\displaystyle{L.H.S=\frac{\tan \theta+\sec \theta-1}{\tan \theta -\sec \theta+1}}$

Now we have ;

$\displaystyle{\sec^2x -\tan^2x = 1}$

putting value of 1 in numerator of L.H.S.

$\displaystyle{L.H.S=\frac{(\tan \theta+\sec \theta)-(\sec^2 \theta - \tan^2 \theta)}{\tan \theta -\sec \theta+1}}\\\\\\\displaystyle{Using \ identity \ a^2-b^2=(a-b)(a+b)}\\\\\\\displaystyle{\implies\frac{(\tan \theta+\sec \theta)-(\sec \theta - \tan \theta)(\sec \theta + \tan \theta)}{\tan \theta -\sec \theta+1}}\\\\\\\displaystyle{Taking \ (\tan \theta+\sec \theta) \ as \ common \ we \ get}$

$\displaystyle{\implies\frac{(\tan \theta+\sec \theta)(1-\sec \theta +\tan \theta)}{\tan \theta -\sec \theta+1}}\\\\\\\displaystyle{After \ cancel \ out \ common \ terms \ we \ get}\\\\\\\displaystyle{\implies(\tan \theta+\sec \theta)}$

Here numerator is in plus , but we have to prove in minus .

Let diving by opposite sign .

$\displaystyle{\implies\left(\tan \theta+\sec \theta\right)\times\left(\dfrac{\sec \theta-\tan \theta}{\sec \theta-\tan \theta }\right)}}\\\\\\\displaystyle{\implies\left(\dfrac{\sec^2 \theta-\tan^2 \theta }{\sec \theta-\tan \theta }\right)}}\\\\\\\displaystyle{Again \ we \ know \ \tan^2 \theta-\sec^2 \theta=1}\\\\\\\displaystyle{\implies\left(\dfrac{1}{\sec \theta-\tan \theta}\right)}}$

L.H.S. = R.H.S.

Hence proved .