Question

cot7 1/2°-cot37 1/2°-cot52 1/2°+cot82 1/2°=​

Answer 1

Step-by-step explanation:

7½° lies in the first quadrant.

Therefore, both sin 7½° and cos 7½° is positive.

For all values of the angle A we know that, sin (α - β) = sin α cos β - cos α sin β.

Therefore, sin 15° = sin (45° - 30°)

= 1√2∙√32 - 1√2∙12

= √32√2 - 12√2

= √3−12√2

Again, for all values of the angle A we know that, cos (α - β) = cos α cos β + sin α sin β.

Therefore, cos 15° = cos (45° - 30°)

cos 15° = cos 45° cos 30° + sin 45° sin 30°

= 1√2∙√32 + 1√2∙12

= √32√2 + 12√2

= √3+12√2

Now cot 7½°

= cos7½°sin7½°

= 2cos7½°∙cos7½°2sin7½°∙cos7½°

= 2cos27½°2sin7½°cos7½°

= 1+cos15°sin15°

= 1+cos(45°−30°)sin(45°−30°)

= 1+√3+12√2√3−12√2

= 2√2+√3+1√3−1

= (2√2+√3+1)(√3+1)(√3−1)(√3+1)

= 2√6+2√2+3+√3+√3+13−1

= 2√6+2√2+2√3+42

= √6 + √2 + √3 + 2

= 2 + √2 + √3 + √6

Answer 2

$\large\underline{\sf{Solution-}}$

Given

$\rm :\longmapsto\:cot7\dfrac{1}{2}\degree -cot37\dfrac{1}{2}\degree - cot52\dfrac{1}{2}\degree + cot82\dfrac{1}{2}\degree$

Let we consider,

$\red{\rm :\longmapsto\:cot7\dfrac{1}{2}\degree + cot82\dfrac{1}{2}\degree}$

can be rewritten as

$\rm \: = \: \: cot7\dfrac{1}{2}\degree + cot \bigg(90\degree - 7\dfrac{1}{2}\degree \bigg)$

We know that

$\boxed{ \rm{ cot(90\degree - x) = tanx}}$

So, using this we get

$\rm \: = \: \: cot7\dfrac{1}{2}\degree + tan7\dfrac{1}{2}\degree$

$\rm \: = \: \: cot\dfrac{15}{2}\degree + tan\dfrac{15}{2}\degree$

Let assume that

$\red{\boxed{ \rm{ \dfrac{15}{2}\degree = x}}}$

$\rm \: = \: \: cotx + tanx$

$\rm \: = \: \: \dfrac{cosx}{sinx} + \dfrac{sinx}{cosx}$

$\rm \: = \: \: \dfrac{ {cos}^{2}x + {sin}^{2} x}{sinxcosx}$

$\rm \: = \: \: \dfrac{1}{sinxcosx}$

$\rm \: = \: \: \dfrac{2}{2sinxcosx}$

$\rm \: = \: \: \dfrac{2}{sin2x}$

$\rm \: = \: \: \dfrac{2}{sin15\degree }$

$\rm \: = \: \: \dfrac{2}{sin(45\degree - 30\degree ) }$

$\rm \: = \: \: \dfrac{2}{sin45\degree cos30\degree - sin30\degree cos45\degree }$

$\rm \: = \: \: \dfrac{2}{\dfrac{1}{ \sqrt{2} } \times \dfrac{ \sqrt{3} }{2} - \dfrac{1}{2} \times \dfrac{ 1 }{2} }$

$\rm \: = \: \: \dfrac{4 \sqrt{2} }{ \sqrt{3} - 1 }$

So, we get

$\boxed{ \bf{ cot7\dfrac{1}{2}\degree + cot87\dfrac{1}{2}\degree = \dfrac{4 \sqrt{2} }{ \sqrt{3} - 1 } }}$

Now, Consider,

$\red{\rm :\longmapsto\:cot37\dfrac{1}{2}\degree + cot52\dfrac{1}{2}\degree}$

can be rewritten as

$\rm \: = \: \: cot37\dfrac{1}{2}\degree + cot \bigg(90\degree - 37\dfrac{1}{2}\degree \bigg)$

Now, we know that

$\boxed{ \rm{ cot(90\degree - x) = tanx}}$

So, using this

$\rm \: = \: \: cot37\dfrac{1}{2}\degree + tan37\dfrac{1}{2}\degree$

$\rm \: = \: \: cot\dfrac{75}{2}\degree + tan\dfrac{75}{2}\degree$

Let we assume that

$\red{\boxed{ \rm{ \dfrac{75}{2}\degree = x}}}$

So, we get

$\rm \: = \: \: cotx + tanx$

$\rm \: = \: \: \dfrac{cosx}{sinx} + \dfrac{sinx}{cosx}$

$\rm \: = \: \: \dfrac{ {cos}^{2}x + {sin}^{2} x}{sinxcosx}$

$\rm \: = \: \: \dfrac{1}{sinxcosx}$

$\rm \: = \: \: \dfrac{2}{2sinxcosx}$

$\rm \: = \: \: \dfrac{2}{sin2x}$

$\rm \: = \: \: \dfrac{2}{sin75\degree }$

$\rm \: = \: \: \dfrac{2}{sin(45\degree + 30\degree )}$

$\rm \: = \: \: \dfrac{2}{sin45\degree cos30\degree + sin30\degree cos45\degree }$

$\rm \: = \: \: \dfrac{2}{\dfrac{1}{ \sqrt{2} } \times \dfrac{ \sqrt{3} }{2} + \dfrac{1}{2} \times \dfrac{ 1 }{2} }$

$\rm \: = \: \: \dfrac{4 \sqrt{2} }{ \sqrt{3} + 1 }$

Hence, we get

$\boxed{ \bf{ cot37\dfrac{1}{2}\degree + cot52\dfrac{1}{2}\degree = \dfrac{4 \sqrt{2} }{ \sqrt{3} + 1 } }}$

Now, Consider

$\rm :\longmapsto\:cot7\dfrac{1}{2}\degree -cot37\dfrac{1}{2}\degree - cot52\dfrac{1}{2}\degree + cot82\dfrac{1}{2}\degree$

can be re-arranged as

$\rm \: = \: \: cot7\dfrac{1}{2}\degree + cot82\dfrac{1}{2}\degree - cot37\dfrac{1}{2}\degree - cot52\dfrac{1}{2}\degree$

$\rm \: = \: \: \bigg(cot7\dfrac{1}{2}\degree + cot82\dfrac{1}{2}\degree \bigg) - \bigg(cot37\dfrac{1}{2}\degree + cot52\dfrac{1}{2}\degree \bigg)$

$\rm \: = \: \: \dfrac{4 \sqrt{2} }{ \sqrt{3} - 1} - \dfrac{4 \sqrt{2} }{ \sqrt{3} + 1}$

$\rm \: = \: \: 4 \sqrt{2}\bigg(\dfrac{1}{ \sqrt{3} - 1} - \dfrac{1}{ \sqrt{3} + 1} \bigg)$

$\rm \: = \: \: 4 \sqrt{2}\bigg(\dfrac{ \sqrt{3} + 1 - \sqrt{3} + 1 }{ (\sqrt{3} - 1)( \sqrt{3} + 1)}\bigg)$

$\rm \: = \: \: 4 \sqrt{2}\bigg(\dfrac{ 2}{ {( \sqrt{3} )}^{2} - {1}^{2} } \bigg)$

$\rm \: = \: \: 4 \sqrt{2}\bigg(\dfrac{ 2}{ 3 - 1 } \bigg)$

$\rm \: = \: \: 4 \sqrt{2}\bigg(\dfrac{ 2}{2} \bigg)$

$\rm \: = \: \: 4 \sqrt{2}$

Hence,

$\rm :\longmapsto\:cot7\dfrac{1}{2}\degree -cot37\dfrac{1}{2}\degree - cot52\dfrac{1}{2}\degree + cot82\dfrac{1}{2}\degree = 4 \sqrt{2}$