Question

If cos θ = 0.6, show that (5 sin θ − 3 tan θ) = 0.

Answer 1

Let us consider a right $△$ABC right angled at B.
Now, we know that cos $\theta$ = 0.6 = $\frac{BC}{AC}$ = $\frac{3}{5}$


So, if BC = 3k, then AC = 5k, where k is a positive number.
Using Pythagoras theorem, we have:
AC² = AB² + BC²
⇒ AB² = AC² $-$ BC²
⇒ AB² = (5k)² $-$ (3k)² = 25k² $-$ 9k²
⇒ AB² = 16k²
⇒ AB = 4k

Finding out the other T-ratios using their definitions, we get:
sin $\theta$ = $\frac{AB}{AC}=\frac{4k}{5k}=\frac{4}{5}$

tan $\theta$ = $\frac{AB}{BC}=\frac{4k}{3k}=\frac{4}{3}$

Substituting the values in the given expression, we get:
5 sin $\theta$ $-$ 3 tan $\theta$
$$\begin{gathered} \Rightarrow 5\left(\frac{4}{5}\right)-3\left({\displaystyle \frac{4}{3}}\right) \\ \Rightarrow 4-4=0=\mathrm{RHS}\hspace{0.17em} \end{gathered}$$
i.e., LHS = RHS

Hence proved.