Question

(tan60+1/tan60-1 )^2=1+cos30/1-cos30

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Hence proved $(\dfrac{\tan 60+1}{\tan60-1})^2=\dfrac{1+\cos 30}{1-\cos 30}$

Step-by-step explanation:

To prove :  $(\dfrac{\tan 60+1}{\tan60-1})^2=\dfrac{1+\cos 30}{1-\cos 30}$

Proof :

Taking LHS,

$LHS=(\dfrac{\tan 60+1}{\tan60-1})^2$

Using trigonometric values, $\tan 60=\sqrt{3}$

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

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

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

$LHS=\dfrac{4+2\sqrt 3}{4-2\sqrt 3}$

$LHS=\dfrac{2+\sqrt 3}{2-\sqrt 3}$

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

We know, $\cos 30=\frac{\sqrt3}{2}$

$LHS=\dfrac{1+\cos 30}{1-\cos 30}$

$LHS=RHS$

Hence proved.

#Learn more

Show that 1-sin60°/cos60°=tan60°-1/tan60°+1

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