The meaning for "YJ"?
Answers
GIVEN :–
ㅤ
• \:\: \bf \tan( \theta) = \dfrac{2x - k}{k \sqrt{3} }tan(θ)=
k
3
2x−k
ㅤ
• \: \: \bf \tan( \phi) = \dfrac{x \sqrt{3} }{2k - x}tan(ϕ)=
2k−x
x
3
ㅤ
TO PROVE :–
ㅤ
• \: \: \bf | \phi-\theta | = {30}^{\circ}∣ϕ−θ∣=30
∘
ㅤ
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)}}}⟹
tan(x−y)=
1+tan(x)tan(y)
tan(x)−tan(y)
ㅤ
• So that –
ㅤ
\: \: \bf \implies \tan( \theta- \phi) = \dfrac{ \tan( \theta) - \tan( \phi)}{1 + \tan( \theta) \tan( \phi)}⟹tan(θ−ϕ)=
1+tan(θ)tan(ϕ)
tan(θ)−tan(ϕ)
ㅤ
• Now put the values –
ㅤ
\: \: \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)}⟹tan(θ−ϕ)=
1+(
k
3
2x−k
)(
2k−x
x
3
)
k
3
2x−k
−
2k−x
x
3
ㅤ
\: \: \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)}⟹tan(θ−ϕ)=
1+(
k
2x−k
)(
2k−x
x
)
(k
3
)(2k−x)
(2x−k)(2k−x)−(k
3
)(x
3
)
ㅤ
\: \: \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)}⟹tan(θ−ϕ)=
1+(
2k
2
−kx
2x
2
−kx
)
(k
3
)(2k−x)
4kx−2x
2
−2k
2
+kx−3kx
ㅤ
\: \: \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)}⟹tan(θ−ϕ)=
(
2k
2
−kx
2k
2
−kx+2x
2
−kx
)
(k
3
)(2k−x)
2kx−2x
2
−2k
2
ㅤ
\: \: \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)}⟹tan(θ−ϕ)=
(
2k
2
−kx
2k
2
−2kx+2x
2
)
(k
3
)(2k−x)
2kx−2x
2
−2k
2
ㅤ
\: \: \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)}⟹tan(θ−ϕ)=
−(
2k
2
−kx
−2k
2
+2kx−2x
2
)
(k
3
)(2k−x)
2kx−2x
2
−2k
2
ㅤ
\: \: \bf \implies \tan( \theta- \phi) = \dfrac{ \dfrac{1}{(k \sqrt{3})(2k - x)}}{ - \left(\dfrac{1}{2 {k}^{2} - kx } \right)}⟹tan(θ−ϕ)=
−(
2k
2
−kx
1
)
(k
3
)(2k−x)
1
ㅤ
\: \: \bf \implies \tan( \theta- \phi) = - \dfrac{2 {k}^{2} - kx}{(k \sqrt{3})(2k - x)}⟹tan(θ−ϕ)=−
(k
3
)(2k−x)
2k
2
−kx
ㅤ
\: \: \bf \implies \tan( \theta- \phi) = - \dfrac{k \cancel{(2k- x)}}{(k \sqrt{3}) \cancel{(2k - x)}}⟹tan(θ−ϕ)=−
(k
3
)
(2k−x)
k
(2k−x)
ㅤ
\: \: \bf \implies \tan( \theta- \phi) = - \dfrac{ \cancel k}{\cancel k \sqrt{3}}⟹tan(θ−ϕ)=−
k
3
k
ㅤ
\: \: \bf \implies \tan( \theta- \phi) = - \dfrac{1}{\sqrt{3}}⟹tan(θ−ϕ)=−
3
1
ㅤ
Henceproved