Question

Pls answer the question in attachment . Don't spam please .​

Answer 1

Answer:

hey miss good morning have a grateful and joyful day

Step-by-step explanation:

is that much thanks is OK or do you want more?

Answer 2

Step-by-step explanation:

We have,

$\tan \bigg( \frac{\pi}{24} \bigg) + \cot \bigg( \frac{\pi}{24} \bigg)$

$= \frac{ \sin \bigg( \dfrac{\pi}{24} \bigg)}{ \cos \bigg( \dfrac{\pi}{24} \bigg) } + \frac{\cos\bigg( \dfrac{\pi}{24} \bigg) }{ \sin \bigg( \dfrac{\pi}{24} \bigg)}$

$= \frac{ \sin ^{2} \bigg( \dfrac{\pi}{24} \bigg) + \cos^{2} \bigg( \dfrac{\pi}{24} \bigg)}{ \cos \bigg( \dfrac{\pi}{24} \bigg) \sin \bigg( \dfrac{\pi}{24} \bigg)}$

$= \frac{ 1}{ \cos \bigg( \dfrac{\pi}{24} \bigg) \sin \bigg( \dfrac{\pi}{24} \bigg)}$

$= \frac{ 2}{ 2\cos \bigg( \dfrac{\pi}{24} \bigg) \sin \bigg( \dfrac{\pi}{24} \bigg)}$

$= \frac{ 2}{ \sin \bigg(2 \times \dfrac{\pi}{24} \bigg)}$

$= \frac{ 2}{ \sin \bigg( \dfrac{\pi}{12} \bigg)}$

$= \frac{ 2}{ \sin \bigg( \dfrac{\pi}{4} - \dfrac{\pi}{6} \bigg)}$

$= \frac{ 2}{ \sin \bigg( \dfrac{\pi}{4} \bigg) \cos \bigg( \dfrac{\pi}{6} \bigg) - \cos \bigg( \dfrac{\pi}{4} \bigg) \sin \bigg( \dfrac{\pi}{6} \bigg) }$

$= \frac{ 2}{ \dfrac{1}{ \sqrt{2} } . \dfrac{ \sqrt{3} }{2} - \dfrac{1}{ \sqrt{2} } . \dfrac{1}{2} }$

$= \frac{ 2}{ \dfrac{ \sqrt{3} - 1 }{2 \sqrt{2} } }$

$= \frac{ 2.2 \sqrt{2} }{ \sqrt{3} - 1 }$

$= \frac{ 4 \sqrt{2}( \sqrt{3} + 1) }{ ( \sqrt{3} - 1 )( \sqrt{3} + 1) }$

$= \frac{ 4 \sqrt{2}( \sqrt{3} + 1) }{ (3 - 1 ) }$

$= \frac{ 4( \sqrt{2} \sqrt{3} + \sqrt{2} ) }{2 }$

$= 2( \sqrt{6} + \sqrt{2} )$