Question

If cos 9 α = sin α and 9α < 90°, then tan 5 α = ?
(a) $\frac{1}{\sqrt{3}}$
(b) $\sqrt{3}$
(c) 1
(d) 0

Answer 1

(c) 1

$$\begin{gathered} \text{We have:} \\ \text{cos9}\alpha \text{=sin}\alpha \\ =>\text{cos9}\alpha =\text{cos}({90}^0-\alpha )[\because \text{sin}\theta =\text{cos}({90}^0-\theta )] \\ =>9\alpha ={90}^0-\alpha . \\ =>10\alpha ={90}^0 \\ =>\alpha =\frac{9{\cancel{0}}^0}{1\cancel{0}} \\ \therefore \alpha =9^0 \\ \text{Again, tan5}\alpha \text{=tan}{(5\times 9)}^0 \\ {\text{=>tan45}}^0=1 \end{gathered}$$