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Q). Show that: tan 75° + cot 75° = 4.​

Sagot :

SalmonWorldTopper

Step-by-step explanation:

[tex]\large\underline{\sf{Solution-}}[/tex]

Given expression is

[tex]\rm :\longmapsto\:tan75\degree + cot75\degree [/tex]

Consider

[tex]\rm :\longmapsto\:tan75\degree [/tex]

[tex]\rm \:  =  \: tan(45\degree + 30\degree )[/tex]

[tex]\rm \:  =  \: \dfrac{tan45\degree + tan30\degree }{1 - tan45\degree \times tan30\degree } [/tex]

[tex]\rm \:  =  \: \dfrac{1 + \dfrac{1}{ \sqrt{3} } }{1 - 1 \times \dfrac{1}{ \sqrt{3} } } [/tex]

[tex]\rm \:  =  \: \dfrac{1 + \dfrac{1}{ \sqrt{3} } }{1 - \dfrac{1}{ \sqrt{3} } } [/tex]

[tex]\rm \:  =  \: \dfrac{ \sqrt{3} + 1 }{ \sqrt{3} - 1} [/tex]

On rationalizing the denominator, we get

[tex]\rm \:  =  \: \dfrac{ \sqrt{3} + 1 }{ \sqrt{3} - 1} \times \dfrac{ \sqrt{3} + 1 }{ \sqrt{3} + 1 } [/tex]

[tex]\rm \:  =  \: \dfrac{ {( \sqrt{3} + 1)}^{2} }{ {( \sqrt{3}) }^{2} - {(1)}^{2} } [/tex]

[tex]\rm \:  =  \: \dfrac{3 + 1 + 2 \sqrt{3} }{3 - 1} [/tex]

[tex]\rm \:  =  \: \dfrac{4+ 2 \sqrt{3} }{2} [/tex]

[tex]\rm \:  =  \: \dfrac{2(2+ \sqrt{3} )}{2} [/tex]

[tex]\rm \:  =  \: 2 + \sqrt{3} [/tex]

[tex]\rm\implies \:\boxed{\tt{ tan75\degree = 2 + \sqrt{3} \: }}[/tex]

Now,

[tex]\rm :\longmapsto\:cot75\degree [/tex]

[tex]\rm \:  =  \: \dfrac{1}{tan75\degree } [/tex]

[tex]\rm \:  =  \: \dfrac{1}{2 + \sqrt{3} } [/tex]

On rationalizing the denominator, we get

[tex]\rm \:  =  \: \dfrac{1}{2 + \sqrt{3} } \times \dfrac{2 - \sqrt{3} }{2 - \sqrt{3} } [/tex]

[tex]\rm \:  =  \: \dfrac{2 - \sqrt{3} }{ {(2)}^{2} - {( \sqrt{3}) }^{2} } [/tex]

[tex]\rm \:  =  \: \dfrac{2 - \sqrt{3} }{4 - 3} [/tex]

[tex]\rm \:  =  \: 2 - \sqrt{3} [/tex]

[tex]\bf\implies \:\boxed{\tt{ cot75\degree = 2 - \sqrt{3} \: }}[/tex]

Now, Consider

[tex]\rm :\longmapsto\:tan75\degree + cot75\degree [/tex]

[tex]\rm \:  =  \: 2 + \sqrt{3} + 2 - \sqrt{3} [/tex]

[tex]\rm \:  =  \: 4[/tex]

Hence,

[tex]\rm\implies \:\boxed{\tt{ tan75\degree + cot75\degree = 4 \: }}[/tex]

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Alternative Method

[tex]\rm :\longmapsto\:tan75\degree + cot75\degree [/tex]

[tex]\rm \:  =  \: tan75\degree + \dfrac{1}{tan75\degree } [/tex]

[tex]\rm \:  =  \: \dfrac{ {tan}^{2}75\degree + 1}{tan75\degree } [/tex]

[tex]\rm \:  =  \: \dfrac{1}{\dfrac{tan75\degree }{1 + {tan}^{2} 75\degree } } [/tex]

[tex]\rm \:  =  \: \dfrac{2}{\dfrac{2tan75\degree }{1 + {tan}^{2} 75\degree } } [/tex]

We know,

[tex]\rm :\longmapsto\:\boxed{\tt{ \frac{2tanx}{1 + {tan}^{2} x} = sin2x}}[/tex]

[tex]\rm \:  =  \: \dfrac{2}{sin150\degree } [/tex]

[tex]\rm \:  =  \: \dfrac{2}{sin(180\degree - 30\degree )} [/tex]

[tex]\rm \:  =  \: \dfrac{2}{sin30\degree } [/tex]

[tex]\rm \:  =  \: 2 \times 2[/tex]

[tex]\rm \:  =  \: 4[/tex]

Hence,

[tex]\rm\implies \:\boxed{\tt{ tan75\degree + cot75\degree = 4 \: }}[/tex]

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