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Prove that $\frac{\cos 10^{\circ}+\sin 10^{\circ}}{\cos 10^{\circ}-\sin 10^{\circ}}=\tan 55^{\circ}$.
To do:
We have to prove that $\frac{\cos 10^{\circ}+\sin 10^{\circ}}{\cos 10^{\circ}-\sin 10^{\circ}}=\tan 55^{\circ}$.
Solution:
We know that,
$\frac{sin\ A}{cos\ A}=tan\ A$
$tan(A+B)=\frac{tan\ A+tan\ B}{1-tan\ A.tan\ B}$
LHS
$\frac{\cos 10^{\circ}+\sin 10^{\circ}}{\cos 10^{\circ}-\sin 10^{\circ}}$
Dividing numerator and denominator by $cos\ 10^o$, we get,
$\frac{\cos 10^{\circ}+\sin 10^{\circ}}{\cos 10^{\circ}-\sin 10^{\circ}}=\frac{1+tan\ 10^o}{1-tan\ 10^o}$
$=\frac{tan\ 45^o+tan\ 10^o}{1-tan\ 45^o.tan\ 10^o}$ ($tan\ 45^o=1$)
$=tan(45^o+10^o)$
$=tan\ 55^o$
$=$RHS
Hence proved.
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