Question

please answer this question .. please please please please please please please please please please please please please please please please​

Answer 1

GIVEN :–

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• $\:\: \bf \tan( \theta) = \dfrac{2x - k}{k \sqrt{3} }$

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• $\: \: \bf \tan( \phi) = \dfrac{x \sqrt{3} }{2k - x}$

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TO PROVE :–

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• $\: \: \bf | \phi-\theta | = {30}^{\circ}$

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SOLUTION :–

• We know that –

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$\: \: \bf \implies \large{ \boxed{ \bf \tan(x - y) = \dfrac{ \tan(x) - \tan(y)}{1 + \tan(x) \tan(y)}}}$

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• So that –

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$\: \: \bf \implies \tan( \theta- \phi) = \dfrac{ \tan( \theta) - \tan( \phi)}{1 + \tan( \theta) \tan( \phi)}$

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• Now put the values –

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$\: \: \bf \implies \tan( \theta- \phi) = \dfrac{ \dfrac{2x - k}{k \sqrt{3} } - \dfrac{x \sqrt{3} }{2k - x} }{1 + \left(\dfrac{2x - k}{k \sqrt{3} } \right) \left(\dfrac{x \sqrt{3} }{2k - x} \right)}$

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$\: \: \bf \implies \tan( \theta- \phi) = \dfrac{ \dfrac{(2x - k)(2k - x) - (k \sqrt{3})(x \sqrt{3} ) }{(k \sqrt{3})(2k - x)}}{1 + \left(\dfrac{2x - k}{k} \right) \left(\dfrac{x}{2k - x} \right)}$

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$\: \: \bf \implies \tan( \theta- \phi) = \dfrac{ \dfrac{4kx - 2 {x}^{2} - 2 {k}^{2} + kx-3kx}{(k \sqrt{3})(2k - x)}}{1 + \left(\dfrac{2 {x}^{2} - kx}{2 {k}^{2} - kx } \right)}$

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$\: \: \bf \implies \tan( \theta- \phi) = \dfrac{ \dfrac{2kx - 2 {x}^{2} - 2 {k}^{2}}{(k \sqrt{3})(2k - x)}}{ \left(\dfrac{2 {k}^{2} - kx + 2 {x}^{2} - kx}{2 {k}^{2} - kx } \right)}$

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$\: \: \bf \implies \tan( \theta- \phi) = \dfrac{ \dfrac{2kx - 2 {x}^{2} - 2 {k}^{2}}{(k \sqrt{3})(2k - x)}}{ \left(\dfrac{2 {k}^{2} - 2kx + 2 {x}^{2} }{2 {k}^{2} - kx } \right)}$

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$\: \: \bf \implies \tan( \theta- \phi) = \dfrac{ \dfrac{2kx - 2 {x}^{2} - 2 {k}^{2}}{(k \sqrt{3})(2k - x)}}{ - \left(\dfrac{ - 2 {k}^{2} + 2kx - 2 {x}^{2} }{2 {k}^{2} - kx } \right)}$

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$\: \: \bf \implies \tan( \theta- \phi) = \dfrac{ \dfrac{1}{(k \sqrt{3})(2k - x)}}{ - \left(\dfrac{1}{2 {k}^{2} - kx } \right)}$

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$\: \: \bf \implies \tan( \theta- \phi) = - \dfrac{2 {k}^{2} - kx}{(k \sqrt{3})(2k - x)}$

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$\: \: \bf \implies \tan( \theta- \phi) = - \dfrac{k \cancel{(2k- x)}}{(k \sqrt{3}) \cancel{(2k - x)}}$

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$\: \: \bf \implies \tan( \theta- \phi) = - \dfrac{ \cancel k}{\cancel k \sqrt{3}}$

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$\: \: \bf \implies \tan( \theta- \phi) = - \dfrac{1}{\sqrt{3}}$

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$\: \: \bf \implies \tan( \theta- \phi) = - \tan( {30}^{ \circ} )$

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$\: \: \bf \implies \tan( \theta- \phi) = \tan( - {30}^{ \circ} ) \:\:\:\:\:\:\:\:\:[ \: \: \because \: \: \tan ( - \theta ) = - \tan( \theta) ]\:$

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$\: \: \bf \implies ( \theta-\phi) = ( - {30}^{ \circ} )$

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$\: \: \bf \implies | \theta-\phi | = | - {30}^{ \circ} |$

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$\: \: \bf \implies \large{ \boxed{ \bf |\phi- \theta | = {30}^{ \circ}}}$

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$\: \: \: \: \: \bf { \underbrace{ \bf Hence \: \: \: proved}}$

Answer 2

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

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$\:\: \bf \tan( \theta) = \dfrac{2x - k}{k \sqrt{3} }tan(θ)$

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$\: \: \bf \tan( \phi) = \dfrac{x \sqrt{3} }{2k - x}tan(ϕ)$

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

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$\: \: \bf | \phi-\theta | = {30}^{\circ}$

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

${\sf{\underline{\underline{\pink{ We \ know \ that –}}}}}$

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$\: \: \bf \implies \large{ \boxed{ \bf \tan(x - y) = \dfrac{ \tan(x) - \tan(y)}{1 + \tan(x) \tan(y)}}}$

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${\sf{\underline{\underline{\pink{So\ that –}}}}}$

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$\: \: \bf \implies \tan( \theta- \phi) = \dfrac{ \tan( \theta) - \tan( \phi)}{1 + \tan( \theta) \tan( \phi)}$

${\sf{\underline{\underline{\pink{Now \ put \ the \ values –}}}}}$

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$\: \: \bf \implies \tan( \theta- \phi) = \dfrac{ \dfrac{2x - k}{k \sqrt{3} } - \dfrac{x \sqrt{3} }{2k - x} }{1 + \left(\dfrac{2x - k}{k \sqrt{3} } \right) \left(\dfrac{x \sqrt{3} }{2k - x} \right)}$

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$\: \: \bf \implies \tan( \theta- \phi) = \dfrac{ \dfrac{(2x - k)(2k - x) - (k \sqrt{3})(x \sqrt{3} ) }{(k \sqrt{3})(2k - x)}}{1 + \left(\dfrac{2x - k}{k} \right) \left(\dfrac{x}{2k - x} \right)}$

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$\: \: \bf \implies \tan( \theta- \phi) = \dfrac{ \dfrac{4kx - 2 {x}^{2} - 2 {k}^{2} + kx-3kx}{(k \sqrt{3})(2k - x)}}{1 + \left(\dfrac{2 {x}^{2} - kx}{2 {k}^{2} - kx } \right)}$

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$¬\: \: \bf \implies \tan( \theta- \phi) = \dfrac{ \dfrac{2kx - 2 {x}^{2} - 2 {k}^{2}}{(k \sqrt{3})(2k - x)}}{ \left(\dfrac{2 {k}^{2} - kx + 2 {x}^{2} - kx}{2 {k}^{2} - kx } \right)}$

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$\: \: \bf \implies \tan( \theta- \phi) = \dfrac{ \dfrac{2kx - 2 {x}^{2} - 2 {k}^{2}}{(k \sqrt{3})(2k - x)}}{ \left(\dfrac{2 {k}^{2} - 2kx + 2 {x}^{2} }{2 {k}^{2} - kx } \right)}$

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$\: \: \bf \implies \tan( \theta- \phi) = \dfrac{ \dfrac{2kx - 2 {x}^{2} - 2 {k}^{2}}{(k \sqrt{3})(2k - x)}}{ - \left(\dfrac{ - 2 {k}^{2} + 2kx - 2 {x}^{2} }{2 {k}^{2} - kx } \right)}$

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$\: \: \bf \implies \tan( \theta- \phi) = \dfrac{ \dfrac{1}{(k \sqrt{3})(2k - x)}}{ - \left(\dfrac{1}{2 {k}^{2} - kx } \right)}$

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$\: \: \bf \implies \tan( \theta- \phi) = - \dfrac{2 {k}^{2} - kx}{(k \sqrt{3})(2k - x)}$

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$\: \: \bf \implies \tan( \theta- \phi) = - \dfrac{k \cancel{(2k- x)}}{(k \sqrt{3}) \cancel{(2k - x)}}$

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$\: \: \bf \implies \tan( \theta- \phi) = - \dfrac{ \cancel k}{\cancel k \sqrt{3}}$

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$\: \: \bf \implies \tan( \theta- \phi) = - \dfrac{1}{\sqrt{3}}$

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$\: \: \bf \implies \tan( \theta- \phi) = - \tan( {30}^{ \circ} )$

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$\: \: \bf \implies \tan( \theta- \phi) = \tan( - {30}^{ \circ} ) \:\:\:\:\:\:\:\:\:[ \: \: \because \: \: \tan ( - \theta ) = - \tan( \theta) ]\:$

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$\: \: \bf \implies ( \theta-\phi) = ( - {30}^{ \circ} )$

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$\: \: \bf \implies | \theta-\phi | = | - {30}^{ \circ}$

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$\: \: \bf \implies \large{ \boxed{ \bf |\phi- \theta | = {30}^{ \circ}}}$

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$\: \: \: \: \: \bf { \underbrace{ \bf Hence \: \: \: proved}}$