Question

prove that 1+cot teta+cosec teta/1-cot teta +cosec teta=cosec teta+cot teta-1/cosec teta-cot teta+1​

Answer 1

Answer:

see image and mark as brainlist

Answer 2

Given:

$\bold{\frac{1+ \ cot \theta+ \ cosec \theta}{1-\ cot \theta +\ cosec \theta} = \frac{\ cosec \theta+\ cot \theta-1}{\ cosec \theta-\ cot \theta+ 1}}$

To prove:

L.H.S = R.H.S

Solution:

$\Rightarrow \bold{\frac{1+ \ cot \theta+ \ cosec \theta}{1-\ cot \theta +\ cosec \theta} = \frac{\ cosec \theta+\ cot \theta-1}{\ cosec \theta-\ cot \theta+ 1}}$

solve L.H.S part:

$\Rightarrow \frac{1+ \ cot \theta+ \ cosec \theta}{1-\ cot \theta +\ cosec \theta}$

change the above value:

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

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

multiply the value by their conjugate:

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

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

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

$= \frac{-2\cos^2 \theta -2\cos \theta}{(-2 \sin \theta \cos \theta )}$

$= \frac{-2\cos \theta (\cos \theta+1 )}{(-2\sin \theta \cos \theta )}\\\\= \frac{ \cos \theta+1 }{\sin \theta}\\\\= \ cot \theta+ \ cosec \theta$

Solve R.H.S part:

$\Rightarrow \frac{\ cosec \theta+\ cot \theta-1}{\ cosec \theta-\ cot \theta+ 1}}\\\\= \frac{\frac{1}{\sin \theta}+\frac{\cos \theta}{\sin \theta} -1}{\frac{1}{\sin \theta}-\frac{\cos \theta}{\sin \theta} +1}}\\\\= \frac{\frac{1+\cos \theta-\sin \theta}{\sin \theta}}{\frac{1-\cos \theta+\sin \theta}{\sin \theta}}}\\\\= \frac{1+\cos \theta-\sin \theta}{\sin \theta} \times \frac{\sin \theta}{1-\cos \theta+\sin \theta}\\\\=\frac{1+\cos \theta-\sin \theta}{1-\cos \theta+\sin \theta}$

multiply the value by their conjugate:

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

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

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

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

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

The proving is in correct because L.H.S ≠ R.H.S