Question

Find sec 2Theta in exact value

Answer 1

Answer:

$sec(2\theta)=-\frac{5}{4}$

Step-by-step explanation:

we know that

step 1

Find $cos(\theta)$

we know that

$tan^2(\theta)+1=sec^2(\theta)$

we have

$tan(\theta)=3$

substitute

$3^2+1=sec^2(\theta)$

$sec^2(\theta)=10$

$sec(\theta)=\pm\sqrt{10}$

Remember that

Angle theta lie in Quadrant I

so

$sec(\theta)$ is positive

$sec(\theta)=\sqrt{10}$

Remember that

$sec(\theta)=\frac{1}{cos(\theta)}$

therefore

$cos(\theta)=\frac{1}{\sqrt{10}}$

step 2

Find $sin(\theta)$

we know that

$tan(\theta)=\frac{sin(\theta)}{cos(\theta)}$

$sin(\theta)=tan(\theta)cos(\theta)$

we have

$cos(\theta)=\frac{1}{\sqrt{10}}$

$tan(\theta)=3$

substitute

$sin(\theta)=(3)(\frac{1}{\sqrt{10}})$

$sin(\theta)=\frac{3}{\sqrt{10}}$

step 3

Find $cos(2\theta)$

we know that

$cos(2\theta)=2cos^2(\theta)-1$

we have

$cos(\theta)=\frac{1}{\sqrt{10}}$

substitute

$cos(2\theta)=2(\frac{1}{\sqrt{10}})^2-1$

$cos(2\theta)=\frac{1}{5}-1$

$cos(2\theta)=-\frac{4}{5}$

Remember that

$sec(2\theta)=\frac{1}{cos(2\theta)}$

therefore

$sec(2\theta)=-\frac{5}{4}$