Given:
There are given the trigonometric function:
[tex]sec^2\theta cos2\theta=1-tan^2\theta[/tex]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:
[tex]sec^2\theta cos2\theta[/tex]Now,
From the formula of cos function:
[tex]cos2\theta=cos^2\theta-sin^2\theta[/tex]Then,
Use the above formula on the above-left side of the equation:
[tex]sec^2\theta cos2\theta=sec^2\theta(cos^2\theta-sin^2\theta)[/tex]Now,
From the formula of sec function:
[tex]sec^2\theta=\frac{1}{cos^2\theta}[/tex]Then,
Apply the above sec function into the above equation:
[tex]\begin{gathered} sec^2\theta cos2\theta=sec^2\theta(cos^2\theta-s\imaginaryI n^2\theta) \\ =\frac{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}[/tex]Then,
[tex]\frac{(cos^{2}\theta- s\mathrm{\imaginaryI}n^{2}\theta)}{cos^{2}\theta}=\frac{cos^2\theta}{cos^2\theta}-\frac{sin^2\theta}{cos^2\theta}[/tex]Then,
From the formula for tan function:
[tex]\frac{sin^2\theta}{cos^2\theta}=tan^2\theta[/tex]Then,
Apply the above formula into the given result:
So,
[tex]\begin{gathered} \frac{(cos^{2}\theta- s\mathrm{\imaginaryI}n^{2}\theta)}{cos^{2}\theta}=\frac{cos^{2}\theta}{cos^{2}\theta}-\frac{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}[/tex]Final answer:
Hence, the above trigonometric function has been proved.
[tex]sec^2\theta cos2\theta=1-tan^2\theta[/tex]