Question

Prove that:
(i)$\frac{cos{11}^{\circ }+sin{11}^{\circ }}{cos{11}^{\circ }-sin{11}^{\circ }}=tan{56}^{\circ }$
(ii)$\frac{cos{9}^{\circ }+sin{9}^{\circ }}{cos{9}^{\circ }-sin{9}^{\circ }}=tan{54}^{\circ }$
(iii)$\frac{cos{8}^{\circ }-sin{8}^{\circ }}{cos{8}^{\circ }+sin{8}^{\circ }}=tan{37}^{\circ }$

Answer 1

(i)$\frac{cos{11}^{\circ }+sin{11}^{\circ }}{cos{11}^{\circ }-sin{11}^{\circ }}$
Dividing numerator and denominator by $cos{11}^{\circ }$, we get
$=\frac{\frac{cos{11}^{\circ }}{cos{11}^{\circ }}+\frac{sin{11}^{\circ }}{cos{11}^{\circ }}}{\frac{cos{11}^{\circ }}{cos{11}^{\circ }}-\frac{sin{11}^{\circ }}{cos{11}^{\circ }}}$
$=\frac{1+tan{11}^{\circ }}{1-tan{11}^{\circ }}$
$=\frac{tan{45}^{\circ }+tan{11}^{\circ }}{1-tan{45}^{\circ }\times tan{11}^{\circ }}$
$=tan({45}^{\circ }+{11}^{\circ })$
$\therefore \frac{cos{11}^{\circ }+sin{11}^{\circ }}{cos{11}^{\circ }-sin{11}^{\circ }}=tan{56}^{\circ }$
Hence proved.
(ii)$LHS:\frac{cos{9}^{\circ }+sin{9}^{\circ }}{cos{9}^{\circ }-sin{9}^{\circ }}$
$\frac{\frac{cos{9}^{\circ }}{cos{9}^{\circ }}+\frac{sin{9}^{\circ }}{cos{9}^{\circ }}}{\frac{cos{9}^{\circ }}{cos{9}^{\circ }}-\frac{sin{9}^{\circ }}{cos{9}^{\circ }}}$
[Dividing numerator and denominator by $cos{9}^{\circ }$]
$=\frac{1+tan{9}^{\circ }}{1-tan{9}^{\circ }}$
$=\frac{tan{45}^{\circ }+tan{9}^{\circ }}{1-tan{45}^{\circ }\times tan{9}^{\circ }}$
=$tan({45}^{\circ }+9^{\circ })$
=$tan{54}^{\circ }$
=RHS
$\therefore LHS=RHS$
Hence proved.
(iii)$\frac{cos{8}^{\circ }-sin{8}^{\circ }}{cos{8}^{\circ }+sin{8}^{\circ }}$
$\frac{\frac{cos{8}^{\circ }}{cos{8}^{\circ }}-\frac{sin{8}^{\circ }}{cos{8}^{\circ }}}{\frac{cos{8}^{\circ }}{cos{8}^{\circ }}+\frac{sin{8}^{\circ }}{cos{8}^{\circ }}}$
[Dividing numerator and denominator by $cos{8}^{\circ }$]
$=\frac{1-tan{8}^{\circ }}{1+tan{8}^{\circ }}$
$=\frac{tan{45}^{\circ }-tan{8}^{\circ }}{1+tan{45}^{\circ }\times tan{8}^{\circ }}$
$=tan({45}^{\circ }-8^{\circ })$
$=tan{37}^{\circ }$
=RHS
$\therefore LHS=RHS$
Hence proved.