Question

If $\cot \mathrm{\theta }=\frac{15}{8}\mathrm{then}\mathrm{evaluate}\frac{\left(1+\sin \mathrm{\theta }\right)\left(1-\sin \mathrm{\theta }\right)}{\left(1+\cos \mathrm{\theta }\right)\left(1-\cos \mathrm{\theta }\right)}$.

Answer 1

$$\begin{gathered} \mathrm{Given}:\cot \theta =\frac{15}{8} \\ \mathrm{Since},\cot \theta =\frac{B}{P} \\ \Rightarrow P=8\mathrm{and}B=15 \\ \mathrm{Using}\mathrm{Pythagoras}\mathrm{theorem}, \\ {P}^2+B^2=H^2 \\ \Rightarrow 8^2+{15}^2=H^2 \\ \Rightarrow H^2=64+225 \\ \Rightarrow H^2=289 \\ \Rightarrow H=17 \\ \mathrm{Therefore}, \\ \sin \theta =\frac{P}{H}=\frac{8}{17} \\ \cos \theta =\frac{B}{H}=\frac{15}{17} \\ \mathrm{Now}, \\ \frac{\left(1+\sin \theta \right)\left(1-\sin \theta \right)}{\left(1+\cos \theta \right)\left(1-\cos \theta \right)}=\frac{1-{\sin }^2\theta }{1-{\cos }^2\theta } \\ =\frac{{\cos }^2\theta }{{\sin }^2\theta }\left(\because {\sin }^2\theta +{\cos }^2\theta =1\right) \\ ={\cot }^2\theta \\ ={\left(\frac{{\displaystyle 15}}{{\displaystyle 8}}\right)}^2 \\ =\frac{225}{64} \\ \mathrm{Hence},\frac{\left(1+\sin \theta \right)\left(1-\sin \theta \right)}{\left(1+\cos \theta \right)\left(1-\cos \theta \right)}=\frac{225}{64}. \end{gathered}$$