Question

cot x + cosec x = √3 যখন 0<x<360​

Answer 1

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Given :-

cot x + cosec x = √3 when x ∈ (0, 2 π )

To Find :-

x when x ∈ (0, 2 π )

Formula used :-

◇ sin(x - y) = sinx cosy - siny cosx

◇ sinx is positive in Ist and 2nd quadrant.

◇ sin ( π/6 ) = 1/2

◇ cos (  π/6 ) = √3/2

Solution :-

$cotx \: + cosecx \: = \sqrt{3} \\ \frac{cosx}{sinx} + \frac{1}{sinx} = \sqrt{3} \\ \frac{cosx + 1}{sinx} = \sqrt{3} \\ cosx \: + 1 = \sqrt{3} sinx \\ ⟹ \: \sqrt{3} sinx - cosx = 1 \\ \large\bold\red{divide \: both \: sides \: by \: 2} \\ ⟹ \: \frac{ \sqrt{3} }{2} sinx \: - \frac{1}{2} cosx \: = \frac{1}{2} \\ ⟹ \: cos \frac{\pi}{6} sinx \: - sin \frac{\pi}{6} cosx \: = \frac{1}{2} \\ ⟹ \: sin(x - \frac{\pi}{6} ) = \frac{1}{2} \\ ⟹ \: sin(x - \frac{\pi}{6}) = sin \frac{\pi}{6} or \: sin(\pi \: - \frac{\pi}{6} ) \\ x - \frac{\pi}{6} = \frac{\pi}{6} or \: x - \frac{\pi}{6} = \pi \: - \frac{\pi}{6} \\ ⟹ \: x = \frac{\pi}{3} \: or \: x = \pi$

but x =  π is rejected as it didn't satisfy the equation bcz cosec π is not defined

$\huge \fcolorbox{black}{cyan}{Hope it helps U}$