Question

I’m stuck on how to verify number 7 and how to find the possible value for sin theta

Answer 1

Given:

There are given the trigonometric function:

`sec^2\theta cos2\theta=1-tan^2\theta`

Step-by-step explanation:

To verify the above trigonometric function, we need to solve the left side of the equation.

So,

From the left side of the given equation:

`sec^2\theta cos2\theta`

Now,

From the formula of cos function:

`cos2\theta=cos^2\theta-sin^2\theta`

Then,

Use the above formula on the above-left side of the equation:

`sec^2\theta cos2\theta=sec^2\theta(cos^2\theta-sin^2\theta)`

Now,

From the formula of sec function:

`sec^2\theta=(1)/(cos^2\theta)`

Then,

Apply the above sec function into the above equation:

`\begin{gathered} sec^2\theta cos2\theta=sec^2\theta(cos^2\theta-s\imaginaryI n^2\theta) \\ =(1)/(cos^2\theta)(cos^2\theta-s\mathrm{i}n^2\theta) \\ =\frac{(cos^2\theta-s\mathrm{i}n^2\theta)}{cos^2\theta} \end{gathered}`

Then,

`\frac{(cos^(2)\theta- s\mathrm{\imaginaryI}n^(2)\theta)}{cos^(2)\theta}=(cos^2\theta)/(cos^2\theta)-(sin^2\theta)/(cos^2\theta)`

Then,

From the formula for tan function:

`(sin^2\theta)/(cos^2\theta)=tan^2\theta`

Then,

Apply the above formula into the given result:

So,

`\begin{gathered} \frac{(cos^(2)\theta- s\mathrm{\imaginaryI}n^(2)\theta)}{cos^(2)\theta}=(cos^(2)\theta)/(cos^(2)\theta)-(s\imaginaryI n^(2)\theta)/(cos^(2)\theta) \\ =1-\frac{s\mathrm{i}n^2\theta}{cos^2\theta} \\ =1-tan^2\theta \end{gathered}`

Final answer:

Hence, the above trigonometric function has been proved.

`sec^2\theta cos2\theta=1-tan^2\theta`