Question

[(tan60°+1)/(tan60°-1)]^2= 1+cos30°/1-cos30°​

Answer 1

Step-by-step explanation:

LHS=(

tan60

o

−1

tan60

o

+1

)

2

=(3^ −1^3 +1 ) 2

= 4−2 ^3^4+2^3

= 1− 2^3^1+ 2^3

= 1−cos30^o^1+cos30^o =RHS

Answer 2

To Prove :-

$(\frac{tan60+1}{tan60-1})^{2}=(\frac{1+cos30}{1-cos30} )$

Proof :-

LHS = $(\frac{tan60+1}{tan60-1} )^{2}$

= $(\frac{\sqrt{3} +1}{\sqrt{3}-1 } )^{2}$

= $\frac{(\sqrt{3})^{2}+(1)^{2}+2\sqrt{3}}{(\sqrt{3})^{2}+(1)^{2}-2\sqrt{3} }$

= $\frac{3+1+2\sqrt{3} }{3+1-2\sqrt{3} }$

= $\frac{2+\sqrt{3} }{2-\sqrt{3} }$

= $\frac{1+\frac{\sqrt{3} }{2} }{1-\frac{\sqrt{3} }{2}}$

We know that, $cos30=\frac{\sqrt{3} }{2}$

RHS = $\frac{1+cos30}{1-cos30}$

= $\frac{1+\frac{\sqrt{3} }{2} }{1-\frac{\sqrt{3} }{2}}$

LHS = RHS hence, proved.

☆ Hope it Helps ☆

$@NightRead$