Question

Please answer this question ✨✨ It's a four mark Question so answer it briefly ✨✨​

Answer 1

hope it's Help you

you also 10th

same I am also 10th

Answer 2

Question :–

Show that : $\: \: { \bold{ \tan( {75}^{ \circ} ) = 2 + \sqrt{3} }} \: \:$

ANSWER :–

TO PROVE :–

$\\ \: \: \to \: \: { \bold{ \tan( {75}^{ \circ} ) = 2 + \sqrt{3} }} \: \:$

SOLUTION :–

• Let's take L.H.S. –

$\\ \: \: = \: \: { \bold{ \tan( {75}^{ \circ} ) }} \: \:$

• We should write this as –

$\\ \: \: = \: \: { \bold{ \tan( {45}^{ \circ} +{30}^{ \circ} ) }} \: \:$

• Now using identity –

$\\ \: \: \large \implies{ \boxed{ \bold{ \tan(A + B) = \dfrac{ \tan(A) + \tan(B) }{1 - \tan(A) \tan(B) } }}} \: \:$

• So that –

$\\ \: \: { \bold{ = \dfrac{ \tan( {45}^{ \circ}) + \tan({30}^{ \circ}) }{1 - \tan( {45}^{ \circ}) \tan({30}^{ \circ}) } }} \: \:$

• We know that –

$\\ \: \: \longrightarrow \: { \bold{ \tan( {45}^{ \circ} ) = 1 }} \: \:$

$\\ \: \: \longrightarrow \: { \bold{ \tan( {30}^{ \circ} ) = \dfrac{1}{ \sqrt{3} } }} \: \:$

• So that –

$\\ \: \: { \bold{ = \dfrac{ 1+ \dfrac{1}{ \sqrt{3} } }{1 - (1) (\dfrac{1}{ \sqrt{3} } )} }} \: \:$

$\\ \: \: { \bold{ = \dfrac{ 1+ \dfrac{1}{ \sqrt{3} } }{1 - \dfrac{1}{ \sqrt{3} } } }} \: \:$

$\\ \: \: { \bold{ = \dfrac{ \dfrac{ \sqrt{3} + 1}{ \sqrt{3} } }{ \dfrac{ \sqrt{3} - 1 }{ \sqrt{3} } } }} \: \:$

$\\ \: \: { \bold{ = \dfrac{ { \sqrt{3} + 1} }{ { \sqrt{3} - 1 } } }} \: \:$

• Now rationalization –

$\\ \: \: { \bold{ = \dfrac{ { \sqrt{3} + 1} }{ { \sqrt{3} - 1 } } \times \dfrac{ \sqrt{3} + 1 }{ \sqrt{3} + 1} }} \: \:$

$\\ \: \: { \bold{ = \dfrac{ ({ \sqrt{3} + 1} ) {}^{2} }{ {( \sqrt{3}) {}^{2} - 1 } } \: \: \: \: \: \: [ \: \: \because \: \: (a + b)(a - b) = {a}^{2} - {b}^{2} ]}}$

$\\ \: \: { \bold{ = \dfrac{ ({ \sqrt{3} ) {}^{2} +( 1} ) {}^{2} + 2 \sqrt{3} }{ {3- 1 } } \: \: \: \: \: \: [ \: \: \because \: \: (a + b) {}^{2} = {a}^{2} + {b}^{2} + 2ab ]}}$

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

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

$\\ \: \: { \bold{ = \dfrac{ \cancel2( 2+ \sqrt{3} ) }{ { \cancel2} } }}$

$\\ \: \: { \bold{ = 2 + \sqrt{3} }}$

$\\ \: \: { \bold{ = R.H.S. \: \: \: \: \: (Hence \: \: proved) }}$