Question

If tan θ = $\frac{4}{3}$, show that (sin θ + cos θ) = $\frac{7}{5}$.

Answer 1

Let us consider a right $△$ABC, right angled at B and $\angle C=\theta$.
Now, we know that tan $\theta$ = $\frac{AB}{BC}$ = $\frac{4}{3}$.

So, if BC = 3k, then AB = 4k, where k is a positive number.
Using Pythagoras theorem, we have:
AC² = AB² + BC² = (4k)² + (3k)²
⇒ AC² = 16k² + 9k² = 25k²
⇒ AC = 5k
Finding out the values of sin $\theta$ and cos $\theta$ using their definitions, we have:
sin $\theta$ = $\frac{AB}{AC}=\frac{4k}{5k}=\frac{4}{5}$
cos $\theta$ = $\frac{BC}{AC}=\frac{3k}{5k}=\frac{3}{5}$
Substituting these values in the given expression, we get:
(sin $\theta$ + cos $\theta$) = $(\frac{4}{5}+\frac{3}{5})=\left(\frac{7}{5}\right)=\mathrm{RHS}$
i.e., LHS = RHS

Hence proved.