Question

please answer step by step​

Answer 1

$\mathsf{Given :\;\dfrac{{sec}^2\phi - sin^2\phi}{tan^2\phi}}$

$\bigstar\;\;\textsf{We know that : \boxed{\mathtt{{sec}\theta = \dfrac{1}{cos\theta}}}}$

$\mathsf{\implies \dfrac{\bigg(\dfrac{1}{cos\phi}\bigg)^2 - sin^2\phi}{tan^2\phi}}$

$\mathsf{\implies \dfrac{\dfrac{1}{cos^2\phi} - sin^2\phi}{tan^2\phi}}$

$\mathsf{\implies \dfrac{\dfrac{1 - sin^2\phi.cos^2\phi}{cos^2\phi}}{tan^2\phi}}$

$\mathsf{\implies \dfrac{1 - sin^2\phi.cos^2\phi}{cos^2\phi.tan^2\phi}}$

$\bigstar\;\;\textsf{We know that : \boxed{\mathtt{tan\theta = \dfrac{sin\theta}{cos\theta}}}}$

$\mathsf{\implies \dfrac{1 - sin^2\phi.cos^2\phi}{cos^2\phi.\bigg(\dfrac{sin\phi}{cos\phi}\bigg)^2}}$

$\mathsf{\implies \dfrac{1 - sin^2\phi.cos^2\phi}{cos^2\phi.\dfrac{sin^2\phi}{cos^2\phi}}}$

$\mathsf{\implies \dfrac{1 - sin^2\phi.cos^2\phi}{sin^2\phi}}$

$\mathsf{\implies \dfrac{1}{sin^2\phi} - \dfrac{sin^2\phi.cos^2\phi}{sin^2\phi}}$

$\mathsf{\implies \bigg(\dfrac{1}{sin\phi}\bigg)^2 - cos^2\phi}$

$\bigstar\;\;\textsf{We know that : \boxed{\mathtt{co{sec}\theta = \dfrac{1}{sin\theta}}}}$

$\mathsf{\implies co{sec}^2\phi - cos^2\phi}$

Answer 2

Answer:

see the attachment

Step-by-step explanation: