Question

If tan ${35}^{\circ }$= k, then the value of $\frac{tan{145}^{\circ }-tan{125}^{\circ }}{1+tan{145}^{\circ }tan{125}^{\circ }}$=

• A. $\frac{2k}{1-k^2}$
• B. $\frac{2k}{1+k^2}$
• C. $\frac{1-k^2}{2k}$
• D. $\frac{1-k^2}{1+k^2}$

Answer 1

The correct option is C $\frac{1-k^2}{2k}$

$\frac{tan{145}^{\circ }-tan{125}^{\circ }}{1+tan{145}^{\circ }tan{125}^{\circ }}=tan({145}^{\circ }-{125}^{\circ })=tan{20}^{\circ }=\frac{1}{tan{70}^{\circ }}$

$=\frac{1-ta{n}^2{35}^{\circ }}{2tan{35}^{\circ }}=\frac{1-k^2}{2k}$