If Secθ + tanθ = p, 0ᵒ < θ < 90ᵒ, then (p² - 1)/(p² + 1) is equal to:

If Secθ + tanθ = p, 0ᵒ < θ < 90ᵒ, then (p² - 1)/(p² + 1)…

If Secθ + tanθ = p, 0^o < θ < 90^o, then (p² - 1)/(p² + 1) is equal to:

Show answer & explanation

Correct answer: A

Explanation

The correct answer is option 1 i.e Sinθ.

Understanding/Application

Secθ + tanθ = p

sq. both sides -

(Secθ + tanθ)² = p² .....(1

(p² - 1)/(p² + 1)

{(Secθ + tanθ)² - 1}/{(Secθ + tanθ) + 1}

(Sec²θ + tan²θ + 2Secθtanθ - 1)/(Sec²θ + tan²θ + 2Secθtanθ + 1)

{(Sec²θ - 1) + tan²θ + 2Secθtanθ }/{ Sec²θ + 2Secθtanθ + (tan²θ + 1)}

(2tan²θ + 2Secθtanθ)/(2Sec²θ + 2Secθtanθ)

(tan²θ + Secθtanθ)/(Sec²θ + Secθtanθ )

{tanθ(1 + Secθ)}/{Secθ(1 + tanθ)}

(tanθ/Secθ)

Sinθ