Question

Ex 23(A).

(Trigonometrical Rational of Standard Angles)

Q3(v) Prove .

The question above is given from the ICSE book (Selina) IX .

Answer 1

ANSWER :

Considering LHS

$= \frac{ {tan}^{2} 60+ 1 + 2tan \: 60}{ {tan}^{2} 60+ 1 - 2tan \: 60 }$

Substituting the value of tan 60,

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

$= \frac{4 + 2 \sqrt{3} }{4 - 2 \sqrt{3} }$

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

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

--------> eq 1

Considering RHS,

$= \frac{1 + cos \: 30}{1 - cos \: 30} = \frac{ 1 + \frac{ \sqrt{3} }{2} }{1 - \frac{ \sqrt{3} }{2} }$

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

----------> eq 2

Eq 1= eq 2

hence proved....

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