Question

(1+cot theta-cosec theta) (1+ tan theta+sec theta) =2

Answer 1

(1 + cos/sin - 1/sin ) (1+1/cos + sin /cos )

( cos + sin - 1) (cos + sin +1 )
---------------------------------------
cos * sin

(cos+sin)² - 1²
-----------------
cos * sin

1+ 2 cos sin - 1
-----------------
cos * sin

2

Answer 2

Given:

$\bold{(1+\cot \theta - \ coses \theta )(1+\tan \theta + \sec \theta) =2}$

To find:

L.H.S=R.H.S

Solution:

$\Rightarrow (1+\cot \theta - \ coses \theta )(1+\tan \theta + \sec \theta) =2\\\\\Rightarrow (1+ \frac{\cos \theta}{\sin \theta} - \frac{1}{\sin \theta} )(1+\frac{\sin \theta}{\cos \theta} + \frac{1}{\cos\theta}) =2\\\\\Rightarrow ( \frac{\sin \theta+\cos \theta-1}{\sin \theta})(\frac{\cos \theta+\sin \theta+1}{\cos \theta}) =2\\\\ \Rightarrow \frac{\sin \theta+\cos \theta-1}{\sin \theta} \times \frac{\cos \theta+\sin \theta+1}{\cos \theta} =2$

$\Rightarrow \frac{(\sin \theta+\cos \theta)^2-(1)^2}{\sin \theta\cos \theta} =2\\\\\therefore \ (\sin \theta+\cos \theta)^2= \sin^2 \theta +\cos^2 \theta+2\sin \theta\cos\theta \\\\\sin^2 \theta+\cos^2 \theta= 1\\\\\Rightarrow \frac{\sin^2 \theta +\cos^2 \theta+2\sin \theta\cos\theta-1}{\sin \theta\cos \theta} =2\\\\\Rightarrow \frac{1+2\sin \theta\cos\theta-1}{\sin \theta\cos \theta} =2\\\\\Rightarrow \frac{2\sin \theta\cos\theta}{\sin \theta\cos \theta} =2\\\\\Rightarrow 2=2$

$\bold{L.H.S=R.H.S}$