0

HETAV PATEL

Grade 11,

Courses New

mtan(thita+30) = ntan (thita+120 ) then m+n/m-n so how can we solve

mtan(thita+30) = ntan (thita+120 ) then m+n/m-n so how can we solve

Grade:11

2 Answers

Amrit Raj
46 Points
6 years ago
m/n=tan(\Theta +120)/tan(\Theta +30) =>m/n=-cot(\Theta +30)/tan(\Theta +30)m/n=tan(\Theta +120)/tan(\Theta +30) =>(m/n+1)/(m/n-1)=-1-cot^{2}(\Theta +30)/1+cot^{2}(\Theta +30)
 
Vikas TU
14149 Points
6 years ago
 
  • m tan(Ѳ+30) = n tan(Ѳ+120)
  • m/n = tan(Ѳ+120)/ tan(Ѳ+30)
  • m/n = ((tanѲ+ tan120)/(1- tanѲ tan120))/(( tanѲ +tan30)/(1- tanѲ tan30))
  • m/n =(( tanѲ -√3)/(1+√3tanѲ))/((√3tanѲ-1)/(√3- tanѲ))
  • m/n=(( tanѲ -√3)* (√3- tanѲ))/( (1+√3 tanѲ)* (√3tanѲ-1))
  • =((tanѲ -√3)2/(1√3 tanѲ)2)
  • (tan2Ѳ +3-2√3 tanѲ)/(1-3tan2Ѳ)
  • m/n= (tan2Ѳ +3-2√3 tanѲ)/(1-3tan2Ѳ)
Applying componendo and dividend
m+n/m-n = (-2 tan2Ѳ+4-2√3 tanѲ)/(2(1-3tan2Ѳ))
m+n/m-n =(- tan2Ѳ+2 -√3 tanѲ)/( 1-3tan2Ѳ)

Think You Can Provide A Better Answer ?

Other Related Questions on Mechanics

View all Questions »

ASK QUESTION

Get your questions answered by the expert for free

Answer and earn gift vouchers

Win Gift vouchers upto Rs 500/-

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Click Here Know More

Complete Self Study Package designed by Industry Leading Experts

Click Here Know More

Live 1-1 coding classes to unleash the Creator in your Child

Click Here Know More

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Click Here Know More