If sin theta = b/a... Then find whole under root (a-b/a+b)+whole under root (a+b/a-b) ?
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√(a-b)/a+b)+ √(a+b)/(a-b)
=√(a-b)²+(a+b)²/√(a²-b²)
=(a+b-2√ab+a+b+2√ab) /√(a²-b²)
=2(a+b)/√(a²-b²)
On squaring,we get
4(a+b)(a+b)/(a+b)(a-b)
=4(a+b)/(a-b)
=(4a/a-b)+(4b/a-b)
put b=a.sin0 note:0=theta
=4a(1+sin0)/a(1-sin0)
=(4+4sin0)/(1-sin0)
=√(a-b)²+(a+b)²/√(a²-b²)
=(a+b-2√ab+a+b+2√ab) /√(a²-b²)
=2(a+b)/√(a²-b²)
On squaring,we get
4(a+b)(a+b)/(a+b)(a-b)
=4(a+b)/(a-b)
=(4a/a-b)+(4b/a-b)
put b=a.sin0 note:0=theta
=4a(1+sin0)/a(1-sin0)
=(4+4sin0)/(1-sin0)
Answered by
7
sin x = b/a . I am gonna tell you a beautiful transformation of such fractions .
Let us consider only ( a - b ) / ( a + b ) . Now divide both the numerator and denominator by a .
1/a ( a-b ) / 1/a ( a + b ) = ( 1 - b/a ) / ( 1 + b/a )
Saw how simple that was !
Similarily ( a + b ) / ( a - b ) = ( 1 + b/a ) / ( 1 - b/a )
So our required expression would be
sqrt ( 1 - b/a ) / ( 1 + b/a ) + sqrt ( 1 + b/a ) / ( 1 - b/a )
Replace with b/a with sin x
sqrt ( 1 - sin x ) / ( 1 + sin x ) + sqrt ( 1 + sinx ) / ( 1 - sin x )
Do fraction addition
= ( 1 - sin x ) + ( 1 + sin x ) / sqrt ( 1 - sin^2 x )
= 2 / sqrt ( cos^2 x ) = 2/|cos x| = 2|sec x |
Note* : Some books tend to ignore | | .
Let us consider only ( a - b ) / ( a + b ) . Now divide both the numerator and denominator by a .
1/a ( a-b ) / 1/a ( a + b ) = ( 1 - b/a ) / ( 1 + b/a )
Saw how simple that was !
Similarily ( a + b ) / ( a - b ) = ( 1 + b/a ) / ( 1 - b/a )
So our required expression would be
sqrt ( 1 - b/a ) / ( 1 + b/a ) + sqrt ( 1 + b/a ) / ( 1 - b/a )
Replace with b/a with sin x
sqrt ( 1 - sin x ) / ( 1 + sin x ) + sqrt ( 1 + sinx ) / ( 1 - sin x )
Do fraction addition
= ( 1 - sin x ) + ( 1 + sin x ) / sqrt ( 1 - sin^2 x )
= 2 / sqrt ( cos^2 x ) = 2/|cos x| = 2|sec x |
Note* : Some books tend to ignore | | .
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