Find the value of tan75° and cot75° (with proof)

1.	Find the value of tan75° and cot75° (with proof)
Answer» ONG>Step-by-step explanation: To FIND:- The values of tan75° and cot75° Solution:- The value of tan75° = tan(45° + 30°) It is in the form → tan(A + B) • A = 45° • B = 30° $\boxed{ \leadsto\sf tan(A + B) = \dfrac{tanA + tanB}{1 - tanA.tanB} }$ $\sf = \dfrac{tan45 + tan30}{1 - tan45 \times tan30}$ $\sf = \dfrac{ \dfrac{ \sqrt{3} + 1}{ \cancel{\sqrt{3}} } }{ \dfrac{ \sqrt{3} - 1 }{ \cancel{\sqrt{3}} } } = \dfrac{ \sqrt{3} + 1}{ \sqrt{3} - 1}$ By RATIONALISING the denominator:- $\sf = \dfrac{ \sqrt{3} + 1}{ \sqrt{3} - 1} \times \dfrac{ \sqrt{3} + 1 }{ \sqrt{3} + 1 }$ $\sf = \dfrac{ (\sqrt{3}+1)^{2} }{ (\sqrt{3})^{2} - {(1)}^{2} }$ $\sf = \dfrac{3 + 1 + 2 \sqrt{3} }{ 3 - 1}$ $\sf = \dfrac{4+ 2 \sqrt{3} }{ 2} = \dfrac{2(2 + \sqrt{3}) }{2}$ $\sf = 2 + \sqrt{3}$ $\dashrightarrow\underline{\boxed{\bf tan75^{\circ}= 2 + \sqrt{3} }}$ ____________________________ The value of cot75° = cot(45° + 30°) It is in the form → cot(A + B) • A = 45° • B = 30° $\boxed{ \leadsto\sf cot(A + B) = \dfrac{cotA.cotB - 1}{ cotA + cotB} }$ $\sf = \dfrac{cot45\times cot30 - 1 }{ cot45 + cot30}$ $\sf = \dfrac{ 1. \sqrt{3} - 1}{1 + \sqrt{3}} = \dfrac{ \sqrt{3} - 1 }{ \sqrt{3} + 1 }$ By rationalising the denominator:- $\sf = \dfrac{ \sqrt{3} - 1}{ \sqrt{3} + 1} \times \dfrac{ \sqrt{3} - 1 }{ \sqrt{3} - 1 }$ $\sf = \dfrac{ (\sqrt{3}-1)^{2} }{ (\sqrt{3})^{2} - {(1)}^{2} }$ $\sf = \dfrac{3 + 1 - 2 \sqrt{3} }{ 3 - 1}$ $\sf = \dfrac{4- 2 \sqrt{3} }{ 2} = \dfrac{2(2 - \sqrt{3}) }{2}$ $\sf = 2 - \sqrt{3}$ $\dashrightarrow\underline{\boxed{\bf cot75^{\circ}= 2 - \sqrt{3}}}$

Answer» ONG>Step-by-step explanation: