Given that :
tan θ + sin θ = a → → →eqn(i)
On squaring both sides then
⇛(tan θ + sin θ)² = a²
⇛tan² θ + 2 tan θ sin θ + sin² θ = a²→ → →eqn(ii)
And
tan θ - sin θ = β → → →eqn(iii)
On squaring both sides then
⇛(tan θ - sin θ)² = β²
⇛tan² θ - 2 tan θ sin θ + sin² θ =β²→ → →eqn(iv)
On Subtracting eqn(iv) from eqn(iii)
⇛(tan² θ + 2 tan θ sin θ + sin² θ) -
(tan² θ - 2 tan θ sin θ + sin² θ) = a²- β²
⇛tan² θ + 2 tan θ sin θ + sin² θ -
tan² θ + 2 tan θ sin θ - sin² θ = a²- β²
⇛2 tan θ sin θ +2 tan θ sin θ = a²- β²
⇛4 tan θ sin θ = a²- β² → → →eqn(v)
Now
a = tan θ + sin θ
β = tan θ - sin θ
a β = (tan θ + sin θ)(tan θ - sin θ)
⇛a β = tan² θ - sin²θ
⇛a β = ( sin² θ/ cos² θ) - sin² θ
⇛a β = sin² θ{ (1/cos² θ)-1}
⇛a β = sin² θ{(1- cos² θ)/cos² θ}
⇛a β = sin² θ × sin² θ/ cos² θ
⇛a β = sin² θ tan² θ
⇛a β = ( sin θ tan θ)²
⇛tan θ sin θ = √(a β) → → → eqn(vi)
On Substituting this value in eqn(v)
from eqn(v)& eqn(vi)
⇛a²- β² = 4√(a β)
Hence, proved.
also read similar questions: Simplify the expression using trigonometric identities (csc θ – csc θ · cos^2 θ). options: A) sin^2 θ B) sin θ · tan θ C) sin^3 θ D) sin θ...
https://brainly.com/question/23898460?referrer
