Solution
Given :-
- sec θ + tan θ = p _______(1)
Show that :-
- (p² - 1)/(p² + 1) = sin θ
Step-by-step explanation
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 .
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