Question

1-sin60°/ cos60° = tan60°-1/ tan60°+1 prove it ​

Answer 1

Answer:

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Step-by-step explanation:

HOPE IT HEPLS

Answer 2

Given :-

• $\sf\dfrac{1 - sin 60°}{cos 60°} = \dfrac{tan60°-1}{tan 60° + 1}$

To Prove :-

• $\sf\dfrac{1 - sin 60°}{cos 60°} = \dfrac{tan 60°-1}{tan 60° + 1}$

Formula to be used :-

• $(a-b)^2 = a^2 + b^2 - 2ab$

• $a^2 - b^2 = (a+b)(a-b)$

Notes :-

• $sin 60° = \dfrac{\sqrt{3}}{2}$

• $cos 60° = \dfrac{1}{2}$

• $tan 60° = \sqrt 3$

Solution :-

L.H.S

$\implies\sf\dfrac{1 - sin 60°}{cos 60°}$

Now put the value of sin 60° and cos 60°.

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

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

$\implies\sf\dfrac{2 - \sqrt 3}{\cancel 2}\times\dfrac{\cancel 2}{1}$

$\implies\sf 2 - \sqrt 3$

R.H.S

$\dfrac{tan60°-1}{tan 60° + 1}$

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

( Put the value of tan 60° and rationalize the denominator)

$\implies\sf\dfrac{\sqrt 3^2 + 1^2 - 2\times\sqrt 3 \times 1 }{\sqrt 3 ^2 - 1^2}$

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

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

$\implies\sf\dfrac{\cancel2(2 - \sqrt 3)}{ \cancel 2 }$

$\implies\sf 2 - \sqrt 3$

Now, compare both sides

$\implies\sf 2 - \sqrt 3$ = $\implies\sf 2 - \sqrt 3$

L.H.S = R.H.S

Hence, proved.