Question

Trigonometry class 11.

Write all the steps needed with proper explanation and use of identities.

Answer 1

$\bold{ANSWER :}$

$Now, \frac{{sin}^{2} A - {sin}^{2} B }{ sin A \: cos A - sin B \: cos B}$

$= \frac{2sin(A+B) \: sin (A-B)}{ 2sinA \: cosA - 2sinB \: cosB}$

$= \frac {2sin( A + B)sin(A - B)}{sin2A - sin2B}$

$= \frac{2sin(A + B) \: sin(A - B)}{2cos(A + B) \: sin(A - B)}$

$= \frac{sin(A + B)}{cos(A + B)}$

$=tan(A + B)$

$\implies \boxed{\bold{\frac{{sin}^{2} A - {sin}^{2} B }{ sin A \: cos A - sin B \: cos B} = tan(A + B)}}$

$\text{Hence, proved.}$

Answer 2

$\bold{Hey \: Mate ! }$

Identity used in your question :

$1). \:sin^{2} x = \frac{1-cos2x}{2} \\\\ 2).\: sin\: x - sin\: y = 2cos(\frac{x+y}{2})sin(\frac{x-y}{2})\\\\ 3).\:sin\: 2x= 2sin\: x \:cos\: x \\\\ 4a)\: \& \: 4b)\: \:cos\: x - cos \: y = - 2sin(\frac{x+y}{2})(\frac{y-x}{2})\\\\ or \: = -2\:sin (\frac{x+y}{2})sin(\frac{x-y}{2})$

$\bold{\underline{Solution}}$


$\frac{sin^{2} \: A - sin^{2}\: B }{sin\: A\: cos\: A - sin\: B \: cos\: B} = tan( A\: + \: B )$



Take LHS



$\implies\large{\frac{\frac{1-cos \: 2A}{2} - \frac{1-cos\: 2B}{2}}{\frac{2(sin\: A cos\: A - cos \: B sin\: B }{2}}}$ ....{identity 1}




$\implies\large{\frac{\cancel{1}-cos\: 2A + cos \:2B -\cancel{1}}{2sin\: A\:\:cos\: A - 2sin\: B\:\:cos \: B}}$




$\implies\large{\frac{ -cos\: 2A + cos\: 2B }{ sin \: 2A - sin\: 2 B}}$ ....{identity 3}



$\implies\large{\frac{-(cos\: 2A - cos\: 2B )}{sin \: 2A - sin\: 2 B}}$



Now using identity (2) & (4b) interchange the denominator and numerator

consider,

x = 2A (&) y = 2B



$\implies\large{\frac{-[-2sin(\frac{2A + 2B}{2})sin(\frac{2A - 2B}{2})]}{2(\frac{2A + 2B}{2})sin(\frac{2A-2B}{2})}}$



$\implies\large{\frac{2sin(\frac{2A + 2B}{2})sin(\frac{2A - 2B}{2})}{2(\frac{2A + 2B}{2})sin(\frac{2A-2B}{2})}}$

after substitution next step is



$\implies\large{\frac{sin\: [\frac{2(A+B)}{2}]}{cos\: [\frac{2(A+B)}{2}]}}$



$\implies\large{\frac{sin(A+B)}{cos(A+B)}}\\\\ \implies \large{tan(A+B)}$


$\bold{\boxed{LHS \:= \:RHS}}$

Hence Proved

Thanks