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Physics-

General

Easy

Question

If the resultant of two forces $left parenthesis A plus B right parenthesis$ and $left parenthesis A minus B right parenthesis$ is $square root of A to the power of 2 end exponent plus B to the power of 2 end exponent end root$ , then the angle between these forces is

$\cos^{- 1} [- \frac{(A^{2} - B^{2})}{A^{2} + B^{2}}]$
$\cos^{- 1} [- \frac{(A^{2} + B^{2})}{{(A}^{2} - B^{2})}]$
$\cos^{- 1} [- \frac{A^{2} + B^{2}}{2 (A^{2} - B^{2})}]$
$\cos^{- 1} [- \frac{2 (A^{2} + B^{2})}{A^{2} - B^{2}}]$

The correct answer is: $cos to the power of negative 1 end exponent invisible function application open square brackets negative fraction numerator A to the power of 2 end exponent plus B to the power of 2 end exponent over denominator 2 left parenthesis A to the power of 2 end exponent minus B to the power of 2 end exponent right parenthesis end fraction close square brackets$

Here, $P equals open parentheses A plus B close parentheses comma blank Q equals open parentheses A minus B close parentheses comma$
$R equals square root of A to the power of 2 end exponent plus B to the power of 2 end exponent end root semicolon$
cos $theta equals fraction numerator R to the power of 2 end exponent minus P to the power of 2 end exponent minus Q to the power of 2 end exponent over denominator 2 P Q end fraction$
$equals fraction numerator open parentheses A to the power of 2 end exponent plus B to the power of 2 end exponent close parentheses minus open parentheses A plus B close parentheses to the power of 2 end exponent minus open parentheses A minus B close parentheses to the power of 2 end exponent over denominator 2 open parentheses A plus B close parentheses left parenthesis A minus B right parenthesis end fraction$
$equals negative open square brackets fraction numerator A to the power of 2 end exponent plus B to the power of 2 end exponent over denominator 2 open parentheses A to the power of 2 end exponent minus B to the power of 2 end exponent close parentheses end fraction close square brackets$
$therefore blank theta equals cos to the power of negative 1 end exponent invisible function application open square brackets negative fraction numerator A to the power of 2 end exponent plus B to the power of 2 end exponent over denominator table row cell 2 left parenthesis A to the power of 2 end exponent minus B to the power of 2 end exponent right parenthesis end cell row cell end cell end table end fraction close square brackets$

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