Question

show that
(1- sin 60⁰)/(cos60⁰)=
(tan60⁰-1)/tan60⁰+1)​

Answer 1

Question:-

$\sf{To\: show\: that\: \dfrac{(1 - Sin60^\circ)}{Cos60^\circ} = \dfrac{Tan60^\circ - 1}{Tan60^\circ+1}}$

Solution:-

LHS,

$\sf{\dfrac{1-Sin^\circ}{Cos60^\circ}}$

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

= $\sf{\dfrac{2-\sqrt{3}}{2}\times 2}$

= $\sf{2 - \sqrt{3}}$

RHS,

$\sf{\dfrac{Tan60^\circ -1 }{Tan60^\circ + 1}}$

= $\sf{\dfrac{\sqrt{3}-1}{\sqrt{3}+1}}$

By Rationalizing the denominator

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

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

= $\sf{\dfrac{{(\sqrt{3})}^{2} + {(1)}^{2} - 2\times1\times\sqrt{3}}{3 - 1}}$

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

= $\sf{\dfrac{4-2\sqrt{3}}{2}}$

Taking 2 as common

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

= $\sf{2 - \sqrt{3}}$

$\sf{<u>\therefore LHS = RHS</u><u>\</u><u>:</u><u>(</u><u>Proved</u><u>)</u>}$