Question

If cot θ = $\frac{15}{8}$, show then evaluate $\frac{(2+2\mathrm{sin\theta })(1-\mathrm{sin\theta })}{(1+\mathrm{cos\theta })(2-2\mathrm{cos\theta })}.$

Answer 1

Let us consider a right $△$ABC, right angled at B.
Now, we know that cot $\theta$ =$\frac{\mathrm{Base}}{\mathrm{Perpendicular}}$ =$\frac{BC}{AB}$ = $\frac{15}{8}$.

Using Pythagoras theorem, we have:
AC² = AB² + BC² = (8k)² + (15k)²
⇒ AC² = 64k² + 225k² = 289k²
⇒ AC = 17k

Finding out the value of sin $\theta$ and cos $\theta$ using their definitions, we have:
sin $\theta$ = $\frac{AB}{AC}=\frac{8k}{17k}=\frac{8}{17}$
cos $\theta$ = $\frac{BC}{AC}=\frac{15k}{17k}=\frac{15}{17}$

Substituting these values in the given expression, we get:
$$\begin{gathered} \frac{\mathit{(}\mathit{2}\mathit{}\mathit{+}\mathit{}\mathit{2}sin\mathit{}\theta \mathit{)}\mathit{(}\mathit{1}\mathit{}\mathit{-}sin\mathit{}\theta \mathit{)}}{\mathit{(}\mathit{1}\mathit{}\mathit{+}\mathit{}cos\mathit{}\theta \mathit{)}\mathit{(}\mathit{2}\mathit{}\mathit{-}\mathit{2}cos\mathit{}\theta \mathit{)}} \\ =\frac{(2+2\times {\displaystyle \frac{8}{17}})(1-{\displaystyle \frac{8}{17}})}{(1+{\displaystyle \frac{15}{17}})(2-2\times {\displaystyle \frac{15}{17}})} \\ =\frac{(2+{\displaystyle \frac{16}{17}})(1-{\displaystyle \frac{8}{17}})}{(1+{\displaystyle \frac{15}{17}})(2-{\displaystyle \frac{30}{17}})} \\ =\frac{\left({\displaystyle \frac{34+16}{17}}\right)\left({\displaystyle \frac{9}{17}}\right)}{\left({\displaystyle \frac{32}{17}}\right)\left({\displaystyle \frac{34-30}{17}}\right)} \\ =\frac{{\displaystyle \frac{50}{17}}\times {\displaystyle \frac{9}{17}}}{{\displaystyle \frac{32}{17}}\times {\displaystyle \frac{4}{17}}}=\frac{225}{64} \end{gathered}$$