Question

Prove the following

$tan(195 \degree) = 2 - \sqrt{3}$


Please provide step by step solution ​

Answer 1

Answer:

Value of tan195 degrees is 2√3+2.

Step-by-step explanation:

Tan195 can be written as tan60+tan60+tan60+tan15. The value of tan60 is √3 and tan15 is 2-√3. So we get

√3+√3+√3+2-√3=2√3+2

hope it helps pls mark me the branliest and thank me

Answer 2

$\large\underline{\sf{Solution-}}$

Given Trigonometric function is

$\rm :\longmapsto\:tan(195 \degree)$

$\rm \:  =  \: tan(90\degree \times 2 + 15\degree)$

We know,

$\boxed{\tt{ tan(180\degree + x) = tanx \: }}$

So, using this identity, we get

$\rm \:  =  \: tan15\degree$

$\rm \:  =  \: tan(45\degree - 30\degree)$

We know,

$\boxed{\tt{ tan(x - y) = \dfrac{tanx - tany}{1 + tanx \: tany}}}$

So, using this identity, we get

$\rm \:  =  \: \dfrac{tan45\degree - tan30\degree}{1 + tan45\degree \: tan30\degree}$

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

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

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

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

$\rm \:  =  \: \dfrac{ {( \sqrt{3} - 1) }^{2} }{ {( \sqrt{3}) }^{2} - {1}^{2} }$

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

$\rm \:  =  \: \dfrac{4 - 2 \sqrt{3} }{2}$

$\rm \:  =  \: \dfrac{2(2 - \sqrt{3})}{2}$

$\rm \:  =  \: 2 - \sqrt{3}$

Hence,

$\rm\implies \:\boxed{\tt{ tan195\degree = 2 - \sqrt{3} \: }}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

LEARN MORE

$\boxed{\tt{ sin(x + y) = sinx \: cosy \: + \: siny \: cosx \: }}$

$\boxed{\tt{ sin(x - y) = sinx \: cosy \: - \: siny \: cosx \: }}$

$\boxed{\tt{ cos(x + y) = cosx \: cosy \: - \: sinx \: siny \: }}$

$\boxed{\tt{ cos(x - y) = cosx \: cosy \: + \: sinx \: siny \: }}$

$\boxed{\tt{ tan(x + y) = \frac{tanx + tany}{1 - tanx \: tany} \: }}$

$\boxed{\tt{ cot(x + y) = \frac{cotx \: coty \: - \: 1}{coty \: + \: cotx}}}$

$\boxed{\tt{ cot(x - y) = \frac{cotx \: coty \: + \: 1}{coty \: - \: cotx}}}$

$\boxed{\tt{ {sin}^{2}x - {sin}^{2}y = sin(x + y) \: sin(x - y) \: }}$

$\boxed{\tt{ {cos}^{2}x - {sin}^{2}y = cos(x + y) \: cos(x - y) \: }}$