Question

Find the value of x ( 0< x< π/2 ) satisfying i) cos x / cosecx+1 + cosx / cosecx - 1 =2

Answer 1

GIVEN :–

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$\bf \implies \dfrac{ \cos(x) }{cosec(x) + 1} + \dfrac{ \cos(x) }{cosec(x) - 1} = 2$

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TO FIND :–

• Value of x (0 < x < π/2) = ?

SOLUTION :–

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$\bf \implies \cos(x) \left[\dfrac{1}{cosec(x) + 1} + \dfrac{1}{cosec(x) - 1} \right]= 2$

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$\bf \implies \cos(x) \left[ \dfrac{cosec(x) - 1 +cosec(x) + 1 }{cosec^{2} (x) - 1} \right]= 2$

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$\bf \implies \cos(x) \left[ \dfrac{cosec(x) +cosec(x) }{cosec^{2} (x) - 1} \right]= 2$

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$\bf \implies \cos(x) \left[ \dfrac{2cosec(x)}{cosec^{2} (x) - 1} \right]= 2$

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$\bf \implies \cos(x) \left[ \dfrac{cosec(x)}{cosec^{2} (x) - 1} \right]=1$

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$\bf \implies \cos(x) \left[ \dfrac{cosec(x)}{ \cot^{2} (x)} \right]=1$

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$\bf \implies \cos(x) \left[ \dfrac{ \dfrac{1}{ \sin(x)}}{ \dfrac{ \cos^{2} (x) }{{ \sin}^{2}(x) }} \right]=1$

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$\bf \implies \cos(x) \left[ \dfrac{ \dfrac{1}{1}}{ \dfrac{ \cos^{2} (x) }{{ \sin}(x) }} \right]=1$

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$\bf \implies \cos(x) \left[ \dfrac{ \sin(x) }{ \cos^{2} (x) } \right]=1$

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$\bf \implies \left[ \dfrac{ \sin(x) }{ \cos(x) } \right]=1$

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$\bf \implies \tan(x) =1$

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$\bf \implies \tan(x) = \tan \left( \dfrac{\pi}{4} \right)$

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$\bf \implies \large{ \boxed{ \bf x = \dfrac{\pi}{4} \left(0 < x < \dfrac{\pi}{2} \right)}}$

Answer 2

${\sf{\underline{\underline{\pink{GIVEN :–}}}}}$

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$\bf \implies \dfrac{ \cos(x) }{cosec(x) + 1} + \dfrac{ \cos(x) }{cosec(x) - 1} = 2$

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${\sf{\underline{\underline{\pink{TO\ FIND :–}}}}}$

• Value of x (0 < x < π/2) = ?

${\sf{\underline{\underline{\pink{SOLUTION :–}}}}}$

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$\bf \implies \cos(x) \left[\dfrac{1}{cosec(x) + 1} + \dfrac{1}{cosec(x) - 1} \right]= 2$

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$\bf \implies \cos(x) \left[ \dfrac{cosec(x) - 1 +cosec(x) + 1 }{cosec^{2} (x) - 1} \right]= 2$

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$\bf \implies \cos(x) \left[ \dfrac{cosec(x) +cosec(x) }{cosec^{2} (x) - 1} \right]= 2$

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$\bf \implies \cos(x) \left[ \dfrac{2cosec(x)}{cosec^{2} (x) - 1} \right]= 2$

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$\bf \implies \cos(x) \left[ \dfrac{cosec(x)}{cosec^{2} (x) - 1} \right]=1$

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$\bf \implies \cos(x) \left[ \dfrac{cosec(x)}{ \cot^{2} (x)} \right]=1$

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$\bf \implies \cos(x) \left[ \dfrac{ \dfrac{1}{ \sin(x)}}{ \dfrac{ \cos^{2} (x) }{{ \sin}^{2}(x) }} \right]=1$

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$\bf \implies \cos(x) \left[ \dfrac{ \dfrac{1}{1}}{ \dfrac{ \cos^{2} (x) }{{ \sin}(x) }} \right]=1$

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$\bf \implies \cos(x) \left[ \dfrac{ \sin(x) }{ \cos^{2} (x) } \right]=1$

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$\bf \implies \left[ \dfrac{ \sin(x) }{ \cos(x) } \right]=1$

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$\bf \implies \tan(x) =1$

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$\bf \implies \tan(x) = \tan \left( \dfrac{\pi}{4} \right)⟹tan(x)$

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$\bf \implies \large{ \boxed{ \bf x = \dfrac{\pi}{4} \left(0 < x < \dfrac{\pi}{2} \right)}}$