Question

prove
$\frac{ \sin(a) - \sin(b) }{ \cos(a) + \cos(b) } + \frac{ \ \cos(a) - \cos(b) }{ \sin(a) + \sin(b) } = 0$
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Answer 1

$\textsf{We may recall...}\\ \\ \\ \large\sin^2(x)+\cos^2(x)=1\\ \\ \\ \textsf{So,}$

$\begin{aligned}&\textsf{LHS}\\ \\ \Longrightarrow\ \ &\frac{\sin(a)-\sin(b)}{\cos(a)+\cos(b)}+\frac{\cos(a)-\cos(b)}{\sin(a)+\sin(b)}\\ \\ \Longrightarrow\ \ &\frac{(\sin(a)-\sin(b))(\sin(a)+\sin(b))+(\cos(a)+\cos(b))(\cos(a)-\cos(b))}{(\cos(a)+\cos(b))(\sin(a)+\sin(b))}\end{aligned}$

$\Longrightarrow\ \ \frac{\sin^2(a)-\sin^2(b)+\cos^2(a)-\cos^2(b)}{(\cos(a)+\cos(b))(\sin(a)+\sin(b))}\\ \\ \Longrightarrow\ \ \frac{\sin^2(a)+\cos^2(a)-\sin^2(b)-\cos^2(b)}{(\cos(a)+\cos(b))(\sin(a)+\sin(b))}\\ \\ \Longrightarrow\ \ \frac{(\sin^2(a)+\cos^2(a))-(\sin^2(b)+\cos^2(b))}{(\cos(a)+\cos(b))(\sin(a)+\sin(b))}$

$\Longrightarrow\ \ \frac{1-1}{(\cos(a)+\cos(b))(\sin(a)+\sin(b))}\\ \\ \Longrightarrow\ \ \frac{0}{(\cos(a)+\cos(b))(\sin(a)+\sin(b))}\\ \\ \Longrightarrow\ \ 0\\ \\ \Longrightarrow\ \ \textsf{RHS}$

$\huge\textsc{\underline{\underline{Hence Proved!!!}}}$

Answer 2

HEre Is Your Ans

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Given :-

$\frac{ \sin(a) - \sin(b) }{ \cos(a) + \cos(b) } + \frac{ \ \cos(a) - \cos(b) }{ \sin(a) + \sin(b) }$

To Prove :-

$\frac{ \sin(a) - \sin(b) }{ \cos(a) + \cos(b) } + \frac{ \ \cos(a) - \cos(b) }{ \sin(a) + \sin(b) } = 0$

Solution :-

$= > \frac{ \sin(a) - \sin(b) }{ \cos(a) + \cos(b) } + \frac{ \ \cos(a) - \cos(b) }{ \sin(a) + \sin(b) } \\ \\ = > \frac{(sin(a) - sin(b)) \: (sin(a) + sin(b)) + (cos(a) - cos(b)) \: \: (cos(a) + cos(b))}{(cos(a) + cos(b)) \: \: (sin(a) + sin(b))} \\ \\ = > \frac{{sin}^{2} (a) - {sin}^{2} (b) + {cos}^{2} (a) - {cos}^{2} (b)}{(cos(a) + cos(b)) \: \: (sin(a) + sin(b))} \\ \\ = > \frac{1 - 1}{(cos(a) + cos(b)) \: \: (sin(a) + sin(b))} \\ \\ = > \frac{0}{(cos(a) + cos(b)) \: \: (sin(a) + sin(b))} \\ \\ = > 0$

Hence , Proved

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