Question

Prove that 1+cos2A/sin2A=cotA and deduce the value of cot15

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Answer 1

TO PROVE :–

$\\ \bf \implies\dfrac{1+ \cos(2A)}{ \sin(2A)}= \cot(A)$

SOLUTION :–

• Let's take L.H.S. –

$\\ \bf \: \: = \: \: \dfrac{1+ \cos(2A)}{ \sin(2A)}$

• We know that –

$\\ \bf \longrightarrow\:\: \orange{\cos(2A)= \: \: 2 \cos ^{2} (A) - 1}$

$\\ \bf \longrightarrow\:\: \orange{\sin(2A)= \: \: 2\sin(A)\cos(A) }$

• So that –

$\\ \bf \: \: = \: \: \dfrac{1+2 \cos ^{2} (A) - 1}{2\sin(A) \cos(A) }$

$\\ \bf \: \: = \: \: \dfrac{2 \cos ^{2} (A)}{2\sin(A) \cos(A) }$

$\\ \bf \: \: = \: \: \dfrac{\cos(A)}{\sin(A) }$

$\\ \bf \: \: = \: \: \cot(A)$

$\\ \bf \: \: = \: \:R.H.S.$

$\\ \large {\green{\: \: \to\: \:{ \boxed{ \underline{ \bf Hence \: \: Proved}}} }}$

$\\\rule{200}{2}$

• Now let's find the value of cot(15°) –

$\\\bf \implies\cot( {15}^{ \circ} ) =\cot( {45}^{ \circ} - {30}^{ \circ} )$

• We know that –

$\\ \implies \red{ \boxed{ \bf\cot( \alpha - \beta ) = \dfrac{ \cot( \alpha ) \cot( \beta ) + 1 }{ \cot( \beta ) - \cot( \alpha ) }}}$

• So that –

$\\\bf \implies\cot( {15}^{ \circ} ) =\dfrac{ \cot({45}^{ \circ} ) \cot({30}^{ \circ}) + 1 }{ \cot({30}^{ \circ}) - \cot({45}^{ \circ} ) }$

$\\\bf \implies\cot( {15}^{ \circ} ) =\dfrac{(1)( \sqrt{3}) + 1 }{( \sqrt{3} - 1) }$

$\\\bf \implies\cot( {15}^{ \circ} ) =\dfrac{ \sqrt{3}+ 1 }{\sqrt{3} - 1 }$

• Now rationalization –

$\\\bf \implies\cot( {15}^{ \circ} ) =\dfrac{ (\sqrt{3}+ 1)(\sqrt{3}+ 1) }{(\sqrt{3} - 1)(\sqrt{3}+ 1) }$

$\\\bf \implies\cot( {15}^{ \circ} ) =\dfrac{ (\sqrt{3}+ 1)^{2}}{(\sqrt{3})^{2} - (1)^{2}}$

$\\\bf \implies\cot( {15}^{ \circ} ) =\dfrac{ (\sqrt{3})^{2} + (1)^{2} + 2 \sqrt{3} }{3-1}$

$\\\bf \implies\cot( {15}^{ \circ} ) =\dfrac{3+1+ 2 \sqrt{3} }{2}$

$\\\bf \implies\cot( {15}^{ \circ} ) =\dfrac{4+ 2 \sqrt{3} }{2}$

$\\\bf \implies\cot( {15}^{ \circ} ) =\dfrac{2(2+ \sqrt{3})}{2}$

$\\\large \implies{ \green{{ \boxed{ \bf\cot( {15}^{ \circ} ) =2 + \sqrt{3}}}}}$

Answer 2

Converting Cot(π/24) into degrees by multiplying the angle by 180/π we can get

Cot(π/24)= cot 7.5

cotA= (1+cos2A)/sin2A

We can write

Cot (15/2)= (1+cos15°)/sin15°……………(1)

Now

cos15°= cos (45°-30°)

=cos45.cos30+sin45.sin30

=1/√2.√3/2+1/√2.1/2

=> cos 15°= (1/4) (√6 + √2)

Similarly sin15° = (1/4) (√6 - √2)

Putting these values in eq. (1)

Cot (15/2)= [1 + (1/4) (√6 + √2)] / [(1/4) (√6 - √2)] 

= (4 + √6 + √2) / (√6 - √2)

On rationalisation 

Cot (15/2)= [(4 + √6 + √2) * (√6 + √2)] / [(√6 - √2) * (√6 + √2)] 

= (4√6 + 6 + 2√3 + 4√2 + 2√3 + 2) / 4 

= (4√2 + 4√3 + 8 + 4√6) / 4 

= √2 + √3 + √4 + √6