Question

please help me friends prove that question as fast as possible it's urgent requirement....​

Answer 1

Question :-

$\frac{1 - \sin(60 {}^{0} ) }{ \cos(60 {}^{0} ) } = \frac{ \tan(60 {}^{0} ) - 1 }{ \tan(60 {}^{0} ) + 1}$

Solution :-

$\frac{1 - \sin(60 {}^{0} ) }{ \cos(60 {}^{0} ) } = \frac{ \tan(60 {}^{0} ) - 1}{ \tan(60 {}^{0} ) + 1} \\ \\ put \: the \: all \: triangular \: values \\ \\ \\ \frac{1 - \frac{ \sqrt{3} }{2} }{ \frac{1}{2} } = \frac{ \sqrt{3} - 1}{ \sqrt{3} + 1 }$

Now solve this

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

Now rationalize Rhs.

$2 - \sqrt{3} = \frac{( \sqrt{3} - 1) {}^{2} }{3 - 1} \\ \\ 2 - \sqrt{3} = \frac{3 + 1 - 2 \sqrt{3} }{2} \\ \\ 2 - \sqrt{3} = \frac{4 - 2 \sqrt{3} }{2} \\ \\ 2 - \sqrt{3} = 2 - \sqrt{3}$

Hence,

Rhs = Lhs (Proved)

Answer 2

${\huge{\bold{\purple{\bigstar{\boxed{\boxed{\bf{Question}}}}}}}}$

Show that :-

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

${\huge{\bold{\purple{\bigstar{\boxed{\boxed{\bf{Solution}}}}}}}}$

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

${\bold{\blue{\boxed{\bf{LHS}}}}}$

• $sin 60 = \dfrac{\sqrt{3}}{2}$

• $cos 60 = \dfrac{1}{2}$

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

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

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

${\bold{\blue{\boxed{\bf{RHS}}}}}$

• $tan 60 = \sqrt 3$

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

• rationalize the denominator

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

• $(a-b)^2 = a^2 + b^2 - 2ab$
• $a^2 - b^2 = (a+b)(a-b)$

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

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

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

$\implies\sf\dfrac{\cancel2(1+ \sqrt 3)}{ \cancel 2 }$

$\implies\sf 1 + \sqrt 3$

${\bold{\blue{\boxed{\bf{Compare}}}}}$

$\implies\sf 1 + \sqrt 3$ = $\implies\sf 1 + \sqrt 3$

LHS = RHS

......Hence proved

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