Plz 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 .

step-by-step explanation:

its 512 because parenthesis and to the 2nd power is 512 so it will be 512

you should talk someone who owns the website about it. you did not deserve to have all that hard work and time taken away. id watch videos of little puppies to cool you down, and maybe some nacho cheese doritos.