Question

If $\mathrm{sin\theta }=\frac{c}{\sqrt{c^2+d^2}}$, where d > 0 then find the values of cos θ and tan θ.

Answer 1

$$\begin{gathered} \mathrm{Given}:\sin \theta =\frac{c}{\sqrt{c^2+d^2}} \\ \mathrm{Since},\sin \theta =\frac{P}{H} \\ \Rightarrow P=c\mathrm{and}H=\sqrt{c^2+d^2} \\ \mathrm{Using}\mathrm{Pythagoras}\mathrm{theorem}, \\ {P}^2+B^2=H^2 \\ \Rightarrow c^2+B^2=c^2+d^2 \\ \Rightarrow B^2=d^2 \\ \Rightarrow B=d \\ \mathrm{Therefore}, \\ \cos \theta =\frac{B}{H}=\frac{d}{\sqrt{c^2+d^2}} \\ \tan \theta =\frac{P}{B}=\frac{c}{d} \\ \mathrm{Hence},\cos \theta =\frac{d}{\sqrt{c^2+d^2}}\mathrm{and}\tan \theta =\frac{c}{d}. \end{gathered}$$