Answer:
![sec(\theta) * cosec(\theta) = (tan^2 (\theta)+ 1)/(tan (\theta)) = tan (\theta)+ (1)/(tan (\theta)) = p + (1)/(p)](https://img.qammunity.org/2021/formulas/mathematics/high-school/gtnfsp1b6a5ko1deslx6h2gp5c2jyczgsh.png)
Explanation:
The given trigonometric relations are
tan(θ) = p
sec(θ)×cosec(θ) = p + 1/p
We note that, when tan(θ) = p, we have;
p + 1/p = tan(θ) + 1/(tan(θ)) = (tan²(θ) + 1)/tan(θ)
By trigonometric ratios, we have;
tan²(θ) + 1 = sec²(θ) =1/cos²(θ) which gives;
(tan²(θ) + 1)/tan(θ) = 1/cos²(θ) × 1/tan(θ) = cos(θ)/sin(θ)×1/cos²(θ)
![(1)/(cos^2(\theta)) * (cos (\theta))/(sin( \theta)) = (1)/(cos(\theta)) * (1)/(sin( \theta)) = sec(\theta) * cosec(\theta)](https://img.qammunity.org/2021/formulas/mathematics/high-school/tiadymxlgopv6ayt5vvpptl5brcb920iia.png)
Therefore;
![If \ tan (\theta) = p \ then \ sec(\theta) * cosec(\theta) = p + (1)/(p)](https://img.qammunity.org/2021/formulas/mathematics/high-school/as2otb4cnetqkhsglpiym91i9csei85km7.png)