# 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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by coolvageesh

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by coolvageesh

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=√(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)

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 | | .