Question

if sinθ=$\frac{1}{3}$ with θ in Q1 find each of the following:1+ $tan^{2}$θ and $sec^{2}$θ

Answer 1

Answer:

$1+\tan^2\theta=\sec^2\theta=\frac{9}{8}$

Explanation:

Let us consider the three sides of a right angled triangle as follows:

Opposite = a
Base = b
Hypotenuse = c

Now

$\sin \theta=\frac{1}{3}$
$\Rightarrow \frac{a}{c}=\frac{1}{3} \Rightarrow c=3a$

In the first question:

$1+\tan^2\theta$
$=1+(\frac{a}{b})^2$
$=\frac{b^2}{b^2}+\frac{a^2}{b^2}=\frac{a^2+b^2}{b^2}$
$=\frac{c^2}{b^2}$

In the second question:

$\sec^2\theta$
$=(\frac{c}{b})^2 = \frac{c^2}{c^2}$

Therefore,

$\frac{c^2}{b^2}=\frac{c^2}{c^2-a^2}$
$=\frac{(3a)^2}{(3a)^2-a^2}=\frac{9a^2}{9a^2-a^2}=\frac{9a^2}{8a^2}$
$=\frac{9}{8}$

$1+\tan^2\theta=\sec^2\theta=\frac{9}{8}$

Hope it helped