Question

Whether the given equation $\frac{sin\Theta -cos\Theta +1}{sin\Theta +cos\Theta -1}=\frac{1}{sin\Theta -tan\Theta }$ is ?

• A. True
• B. False

Answer 1

The correct option is A True

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

L.H.S

$\frac{(\sin \theta -\cos \theta +1)/\cos \theta }{(\sin \theta +\cos \theta -1)/\cos \theta }$

$=\frac{\tan \theta -1+\sec \theta }{\tan \theta +1-\sec \theta }$

$=\frac{\tan \theta -(1-\sec \theta )}{\tan \theta +(1-\sec \theta )}\times \frac{\tan \theta -(1-\sec \theta )}{\tan \theta -(1-\sec \theta )}$

$=\frac{(\tan \theta -(1-\sec \theta ){)}^2}{{\tan }^2\theta -(1-\sec \theta {)}^2}$

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

$=\frac{(2\sec \theta -2)(\sec \theta +\tan \theta )}{2(\sec \theta -1)}=\frac{2(\sec \theta +\tan \theta )}{2}$

$=\frac{1}{(\sec \theta -\tan \theta )}\ne$ R.H.S.

Hence this statement is false.