Question

If tan theta = 3/4, evaluate 1/sin theta+1/cos theta​

Answer 1

Answer:

$\frac{7}{12}$

Step-by-step explanation:

Given:

$tan\theta=\frac{3}{4}$

To find:

$\frac{1}{sin\theta}+\frac{1}{cos\theta}$

Solution:

Let us name the triangle ABC

$tan\theta=\frac{3}{4}$ $($∵$Given)$

$tan\theta=\frac{3k}{4k}$,  where $k$ is some positive real number.

Since, $tan\theta=\frac{Opposite}{Adjacent}$

$tan\theta=\frac{AB}{BC}$

∴ $\frac{AB}{BC} =\frac{3k}{4k}$ $($∵ Things which are equal to the same thing, are equal to one another)

∴ $AB=3k,BC=4k$

By Pythagoras Theorem,

$AC^2=AB^2+BC^2$

$AC^2=(3k)^2+(4k)^2$

$AC^2=9k^2+16k^2$

$AC^2=25k^2$

$AC=5k$

$sin\theta=\frac{Opposite}{Hypotenuse}$

$sin\theta=\frac{AB}{AC}$

$sin\theta=\frac{3k}{5k}$

$sin\theta=\frac{3}{5}$

$cos\theta=\frac{Adjacent}{Hypotenuse}$

$cos\theta=\frac{BC}{AC}$

$cos\theta=\frac{4k}{5k}$

$cos\theta=\frac{4}{5}$

$\frac{1}{sin\theta}+\frac{1}{cos\theta}$

$=\frac{sin\theta+cos\theta}{sin\theta cos\theta}$

⇒ $\frac{\frac{3}{5} +\frac{4}{5} }{\frac{3}{5} \times \frac{4}{5} }$

⇒ $\frac{\frac{3+4}{5} }{\frac{12}{5} }$

⇒ $\frac{\frac{7}{5} }{\frac{12}{5} }$

⇒ $\frac{7}{5} \times \frac{5}{12}$

⇒ $\frac{7}{12}$

∴ $\frac{1}{sin\theta}+\frac{1}{cos\theta}=\frac{7}{12}$