Question:
Given that, sin x + cos x + tan x + cot x + sec x + cosec x = 7. The given equation can be reduced to
दिया गया है कि, \(\sin x + \cos x + \tan x + \cot x + \sec x + \operatorname{cosec} x = 7\) है। दिए गए समीकरण को किस रूप में रूपांतरित किया जा सकता है?
Options:
- \(\sin^2 2x + 44\sin 2x – 36 = 0\)
- \(\sin^2 2x + 22\sin 2x – 18 = 0\)
- \(\sin^2 2x – 44\sin 2x + 36 = 0\)
- \(\sin^2 2x – 22\sin 2x + 18 = 0\)
- \(\sin^2 2x + 44\sin 2x – 36 = 0\)
- \(\sin^2 2x + 22\sin 2x – 18 = 0\)
- \(\sin^2 2x – 44\sin 2x + 36 = 0\)
- \(\sin^2 2x – 22\sin 2x + 18 = 0\)
Correct Answer: c)
Explanation:
Let \(t = \sin x + \cos x\) and \(u = \sin 2x = t^2 – 1\).
\(\sin x + \cos x + \dfrac{\sin x + \cos x}{\sin x \cos x} + \dfrac{1}{\sin x \cos x} = t + \dfrac{t+1}{u/2} = 7\).
Substituting and simplifying leads to the quadratic equation \(\sin^2 2x – 44\sin 2x + 36 = 0\).
माना \(t = \sin x + \cos x\) तथा \(\sin 2x = t^2 – 1\)। मान प्रतिस्थापित कर हल करने पर समीकरण \(\sin^2 2x – 44\sin 2x + 36 = 0\) प्राप्त होता है।