English, asked by kaamyagarg162007, 7 months ago

Ex 23(A).

(Trigonometrical Rational of Standard Angles)

Q3(v) Prove .

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

Attachments:

Answers

Answered by AngelineSudhagar
0

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....

Hope it helps

thank my answer

and mark as brainliest

Similar questions