[2/root6+root2]tan25 +4sin25=[root a+root2]/2 find the value of "a"?
Answers
Answer:
2
+
3
2
6
+
6
+
3
6
2
−
6
+
2
8
3
\begin{gathered}i) \frac{2\sqrt{6}}{\sqrt{2}+\sqrt{3}}\\=\frac{2\sqrt{6}}{\sqrt{3}+\sqrt{2}}\\=\frac{2\sqrt{6}(\sqrt{3}-\sqrt{2})}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}\\=\frac{2\sqrt{6}(\sqrt{3}-\sqrt{2})}{(\sqrt{3})^{2}-(\sqrt{2})^{2}}\\=\frac{2\sqrt{6}(\sqrt{3}-\sqrt{2})}{3-2)}\\=2\sqrt{6}(\sqrt{3}-\sqrt{2})\\=2\sqrt{6}\times \sqrt{3} - 2\sqrt{6}\times \sqrt{2}\\=6\sqrt{2} - 4\sqrt{3} \:---(1)\end{gathered}
i)
2
+
3
2
6
=
3
+
2
2
6
=
(
3
+
2
)(
3
−
2
)
2
6
(
3
−
2
)
=
(
3
)
2
−(
2
)
2
2
6
(
3
−
2
)
=
3−2)
2
6
(
3
−
2
)
=2
6
(
3
−
2
)
=2
6
×
3
−2
6
×
2
=6
2
−4
3
−−−(1)
\begin{gathered}ii) \frac{6\sqrt{2}}{\sqrt{6}+\sqrt{3}}\\=\frac{6\sqrt{2}(\sqrt{6}-\sqrt{3})}{(\sqrt{6}+\sqrt{3})(\sqrt{6}-\sqrt{3})}\\=\frac{6\sqrt{2}(\sqrt{6}-\sqrt{3})}{(\sqrt{6})^{2}-(\sqrt{3})^{2}}\\=\frac{6\sqrt{2}(\sqrt{6}-\sqrt{3})}{6-3)}\\=\frac{6\sqrt{2}(\sqrt{6}-\sqrt{3})}{6-3}\\=\frac{6\sqrt{2}(\sqrt{6}-\sqrt{3})}{3}\\=2\sqrt{2}(\sqrt{6}-\sqrt{3})\\=4\sqrt{3}-2\sqrt{6}\:---(2)\end{gathered}
ii)
6
+
3
6
2
=
(
6
+
3
)(
6
−
3
)
6
2
(
6
−
3
)
=
(
6
)
2
−(
3
)
2
6
2
(
6
−
3
)
=
6−3)
6
2
(
6
−
3
)
=
6−3
6
2
(
6
−
3
)
=
3
6
2
(
6
−
3
)
=2
2
(
6
−
3
)
=4
3
−2
6
−−−(2)
\begin{gathered}iii)\frac{8\sqrt{3}}{\sqrt{6}+\sqrt{2}}\\=\frac{8\sqrt{2}(\sqrt{6}-\sqrt{2})}{(\sqrt{6}+\sqrt{2})(\sqrt{6}-\sqrt{2})}\\=\frac{8\sqrt{2}(\sqrt{6}-\sqrt{2})}{(\sqrt{6})^{2}-(\sqrt{2}))^{2}}\\=\frac{8\sqrt{2}(\sqrt{6}-\sqrt{2})}{6-2}\\=\frac{8\sqrt{2}(\sqrt{6}-\sqrt{2})}{4}\\=2\sqrt{3}(\sqrt{6}-\sqrt{2})\\=6\sqrt{2}-4\:---(3)\end{gathered}
iii)
6
+
2
8
3
=
(
6
+
2
)(
6
−
2
)
8
2
(
6
−
2
)
=
(
6
)
2
−(
2
))
2
8
2
(
6
−
2
)
=
6−2
8
2
(
6
−
2
)
=
4
8
2
(
6
−
2
)
=2
3
(
6
−
2
)
=6
2
−4−−−(3)
/* According to the problem given
\frac{2\sqrt{6}}{\sqrt{2}+\sqrt{3}}+\frac{6\sqrt{2}}{\sqrt{6}+\sqrt{3}}-\frac{8\sqrt{3}}{\sqrt{6}+\sqrt{2}}
2
+
3
2
6
+
6
+
3
6
2
−
6
+
2
8
3
= 6\sqrt{2} - 4\sqrt{3} +4\sqrt{3}-2\sqrt{6}-(6\sqrt{2}-2\sqrt{6})=6
2
−4
3
+4
3
−2
6
−(6
2
−2
6
)
= 6\sqrt{2} - 4\sqrt{3} +4\sqrt{3}-2\sqrt{6}-6\sqrt{2}+2\sqrt{6}=6
2
−4
3
+4
3
−2
6
−6
2
+2
6
= 0=0
Therefore.,
\frac{2\sqrt{6}}{\sqrt{2}+\sqrt{3}}+\frac{6\sqrt{2}}{\sqrt{6}+\sqrt{3}}-\frac{8\sqrt{3}}{\sqrt{6}+\sqrt{2}}
2
+
3
2
6
+
6
+
3
6
2
−
6
+
2
8
3
\green {= 0}=0
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