Math, asked by niteshsingh42, 10 months ago

prove that value of tan 15°​

Answers

Answered by shrutikamble3032
1

Answer:

tan(15) = tan(45 - 30) \\ \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: ...tan (a - b) =  \frac{tana - tanb}{1 +  \tan \: a \times b} \\   \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: =   \frac{tan45 - tan30}{1 + tan \: 45 \:  \times tan \: 30}  \\   \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  = \frac{1 -  \frac{1}{ \sqrt{3} } }{1 +1\times \frac{1}{ \sqrt{3} }  }  \\   \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: =  \frac{ \frac{ \sqrt{3 - 1} }{ \sqrt{3} } }{ \frac{ \sqrt{3 + 1} }{ \sqrt{3} } }  \\   \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  =  \frac{ \sqrt{3 - 1} }{ \sqrt{3} }   \times  \frac{ \sqrt{3} }{ \sqrt{3 + 1 } }  \\  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:   =   \frac{ \sqrt{3 - 1} }{ \sqrt{3 + 1} }

therefore tan (15)=

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

mark me as brainliest !

Answered by Secretgirl123
0

Answer:

tan(45-30)

tan45 - tan30/1 + tan45tan30

(1-1/√3)/(1+1/√3)

√3-1/√3+1

#Secretgirl✌️

Similar questions