Plzzzzz help me............​

Given :-

Show that :-

we Know,

★ sec² θ - tan² θ = 1

★ (a + b )² = a² + b² + 2ab

Then,

Take L.H.S.

= (p² - 1)/(p² + 1)

keep value of p .

= {(sec θ + tan θ)² - 1}/{(sec θ + tan θ)² + 1}

= {(sec² θ + tan² θ + 2sec θ . tan θ) - 1 }/{(sec² θ + tan² θ + 2sec θ . tan θ) + 1}

= ( sec² θ - 1) + ( tan² θ + 2 sec θ . tan θ )}\{(tan² θ + 1) + ( sec ² θ + 2 sec θ . tan θ )

= { tan² θ + tan² θ + 2 sec θ . tan θ }/{(sec² θ + sec²θ + 2sec θ . tan θ]

= (2 tan ² θ + 2 sec θ . tan θ)/(2 sec ² θ + 2 sec θ . tan θ)

= {2tan θ( tan θ + sec θ)}/{2sec θ(sec θ + tan θ) }

= 2 tan θ /2 sec θ

= tan θ/sec θ

= (sin θ/cos θ)/(1/cos θ)

= sin θ

R.H.S.

That's proved .

I would love to but I have insecurities and I dont really know how to get on a zo.om call/go.ogle meet ;w;''