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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Updated on: 10-Oct-2022

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