Question

any one solve this question and give solutions​

Answer 1

Answer:

Step-by-step explanation:

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Answer 2

Proving "both equation is same".

Step-by-step explanation:

$\ Given \ value:\\\\(1+ \cot \theta - \ cosec \theta) (1+ \tan \theta + \sec \theta) = 2\\\\\ Solution:\\\\(1+ \cot \theta - \ cosec \theta) (1+ \tan \theta + \sec \theta) = 2\\\\\ Solve \ L.H.S \ part \\\\ \rightarrow (1+ \frac{\cos \theta }{\sin \theta}- \frac{1}{\sin \theta}) (1+ \frac{\sin \theta}{\cos \theta}+ \frac{1}{\cos \theta}) \\\\ \rightarrow (\frac{\sin \theta+\cos \theta - 1 }{\sin \theta}) (\frac {\cos \theta+\sin \theta+1}{\cos \theta})$

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

$\rightarrow (\frac {(\sin \theta+\cos \theta - 1 )(\cos \theta+\sin \theta+1)}{\sin \theta \cos \theta}) \\\\\ formula: \\\\ \ a^2-b^2 = (a+b) (a-b) \\\\ \rightarrow (\frac {(\sin \theta+\cos \theta - 1)(\sin \theta+\cos \theta+1)}{\sin \theta \cos \theta}) \\\\\rightarrow \frac {(\sin \theta+\cos \theta)^2 -(1)^2}{\sin \theta \cos \theta} \\\\\rightarrow \frac {(\sin^2 \theta+\cos^2 \theta+ 2\sin \theta \cos \theta) - 1}{\sin \theta \cos \theta}$

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

L.H.S = R.H.S

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