cot 7 1/2 degrees - cot 37 1/2 degrees - cot 52 1/2 degrees + cot 82 1/2 degrees?
Answers
Answer:
⟼cot7
2
1
°−cot37
2
1
°−cot52
2
1
°+cot82
2
1
°
Let we consider,
\red{\rm :\longmapsto\:cot7\dfrac{1}{2}\degree + cot82\dfrac{1}{2}\degree}:⟼cot7
2
1
°+cot82
2
1
°
can be rewritten as
\rm \: = \: \: cot7\dfrac{1}{2}\degree + cot \bigg(90\degree - 7\dfrac{1}{2}\degree \bigg)=cot7
2
1
°+cot(90°−7
2
1
°)
We know that
\boxed{ \rm{ cot(90\degree - x) = tanx}}
cot(90°−x)=tanx
So, using this we get
\rm \: = \: \: cot7\dfrac{1}{2}\degree + tan7\dfrac{1}{2}\degree=cot7
2
1
°+tan7
2
1
°
\rm \: = \: \: cot\dfrac{15}{2}\degree + tan\dfrac{15}{2}\degree=cot
2
15
°+tan
2
15
°
Let assume that
\red{\boxed{ \rm{ \dfrac{15}{2}\degree = x}}}
2
15
°=x
\rm \: = \: \: cotx + tanx=cotx+tanx
\rm \: = \: \: \dfrac{cosx}{sinx} + \dfrac{sinx}{cosx}=
sinx
cosx
+
cosx
sinx
\rm \: = \: \: \dfrac{ {cos}^{2}x + {sin}^{2} x}{sinxcosx}=
sinxcosx
cos
2
x+sin
2
x
\rm \: = \: \: \dfrac{1}{sinxcosx}=
sinxcosx
1
\rm \: = \: \: \dfrac{2}{2sinxcosx}=
2sinxcosx
2
\rm \: = \: \: \dfrac{2}{sin2x}=
sin2x
2
\rm \: = \: \: \dfrac{2}{sin15\degree }=
sin15°
2
\rm \: = \: \: \dfrac{2}{sin(45\degree - 30\degree ) }=
sin(45°−30°)
2
\rm \: = \: \: \dfrac{2}{sin45\degree cos30\degree - sin30\degree cos45\degree }=
sin45°cos30°−sin30°cos45°
2
\rm \: = \: \: \dfrac{2}{\dfrac{1}{ \sqrt{2} } \times \dfrac{ \sqrt{3} }{2} - \dfrac{1}{2} \times \dfrac{ 1 }{2} }=
2
1
×
2
3
−
2
1
×
2
1
2
\rm \: = \: \: \dfrac{4 \sqrt{2} }{ \sqrt{3} - 1 }=
3
−1
4
2
So, we get
\boxed{ \bf{ cot7\dfrac{1}{2}\degree + cot87\dfrac{1}{2}\degree = \dfrac{4 \sqrt{2} }{ \sqrt{3} - 1 } }}
cot7
2
1
°+cot87
2
1
°=
3
−1
4
2
Now, Consider,
\red{\rm :\longmapsto\:cot37\dfrac{1}{2}\degree + cot52\dfrac{1}{2}\degree}:⟼cot37
2
1
°+cot52
2
1
°
can be rewritten as
\rm \: = \: \: cot37\dfrac{1}{2}\degree + cot \bigg(90\degree - 37\dfrac{1}{2}\degree \bigg)=cot37
2
1
°+cot(90°−37
2
1
°)
Now, we know that
\boxed{ \rm{ cot(90\degree - x) = tanx}}
cot(90°−x)=tanx
So, using this
\rm \: = \: \: cot37\dfrac{1}{2}\degree + tan37\dfrac{1}{2}\degree=cot37
2
1
°+tan37
2
1
°
\rm \: = \: \: cot\dfrac{75}{2}\degree + tan\dfrac{75}{2}\degree=cot
2
75
°+tan
2
75
°
Let we assume that
\red{\boxed{ \rm{ \dfrac{75}{2}\degree = x}}}
2
75
°=x
So, we get
\rm \: = \: \: cotx + tanx=cotx+tanx
\rm \: = \: \: \dfrac{cosx}{sinx} + \dfrac{sinx}{cosx}=
sinx
cosx
+
cosx
sinx
\rm \: = \: \: \dfrac{ {cos}^{2}x + {sin}^{2} x}{sinxcosx}=
sinxcosx
cos
2
x+sin
2
x
\rm \: = \: \: \dfrac{1}{sinxcosx}=
sinxcosx
1
\rm \: = \: \: \dfrac{2}{2sinxcosx}=
2sinxcosx
2
\rm \: = \: \: \dfrac{2}{sin2x}=
sin2x
2
\rm \: = \: \: \dfrac{2}{sin75\degree }=
sin75°
2
\rm \: = \: \: \dfrac{2}{sin(45\degree + 30\degree )}=
sin(45°+30°)
2
\rm \: = \: \: \dfrac{2}{sin45\degree cos30\degree + sin30\degree cos45\degree }=
sin45°cos30°+sin30°cos45°
2
\rm \: = \: \: \dfrac{2}{\dfrac{1}{ \sqrt{2} } \times \dfrac{ \sqrt{3} }{2} + \dfrac{1}{2} \times \dfrac{ 1 }{2} }=
2
1
×
2
3
+
2
1
×
2
1
2
\rm \: = \: \: \dfrac{4 \sqrt{2} }{ \sqrt{3} + 1 }=
3
+1
4
2
Hence, we get
\boxed{ \bf{ cot37\dfrac{1}{2}\degree + cot52\dfrac{1}{2}\degree = \dfrac{4 \sqrt{2} }{ \sqrt{3} + 1 } }}
cot37
2
1
°+cot52
2
1
°=
3
+1
4
2