Question

If $\tan \theta =\frac{\cos 9^o+\sin 9^o}{\cos 9^o-\sin 9^o}$ then $\theta$ is equal to

• A. ${18}^o$
• B. ${54}^o$
• C. $9^o$
• D. ${36}^o$

Answer 1

Answer: (B)

${54}^o$

$tan\theta =\frac{cos9+sin9}{cos9-sin9}$ dividing by cos 9

$=\frac{1+tan9}{1-tan9}$ $tan45=1$

$=\frac{tan45+tan9}{1-tan45+tan9}$ $tan(A+B)=\frac{tanA+tanB}{1-tanA.TanB}$

$=tan(45+9)$

$tan\theta =tan54$

$\therefore \theta ={54}^{\circ }$