please answer my question number 28 its urgent please fast guys??
Answers
(i)
Given: cosecθ - sinθ = a³
⇒ (1/sinθ) - sinθ = a³
⇒ (1 - sin²θ)/sinθ = a³
⇒ cos²θ/sinθ = a³.
(ii)
Given: secθ - cosθ = b³
⇒ (1/cosθ) - cosθ = b³
⇒ (1 - cos²θ)/cosθ = b³
⇒ sin²θ/cosθ = b³.
LHS:
a²b²(a² + b²)
= a⁴b² + a²b⁴
= {[cos²θ/sinθ]^4/3 * [sin²θ/cosθ]^4/3} + {[cos²θ/sinθ]^2/3 * [sin²θ/cosθ]^4/3
= {[(cosθ)^(8/3)/(cosθ)^(2/3)}] + {[(sinθ)^(8/3)/(sinθ)^2/3}]
= [cosθ]^(8/3 - 2/3) + [sinθ]^(8/3 - 2/3)
= [cosθ]² + [sinθ]²
= 1
= RHS
Hope it helps you!
Step-by-step explanation:
Consider cosec theta - sin theta = a³
⇒ !/sin theta - sin theta = a³
⇒ 1 - sin² theta/sin theta = a³
cos² theta/ sin theta = a³ → (1)
⇒ (cos² theta/sin theta)²/³ = (a³)²/³
⇒ cos⁴/³ theta/sin²/³ theta = a² → (2)
Now consider, sec theta - cos theta = b³
⇒ 1/cos theta - cos theta = b³
⇒ 1 - cos²theta/cos theta = b³
⇒ sin² theta/cos theta = b³ → (3)
⇒ (sin² theta/cos theta)²/³ = (b³)²/³
⇒ sin⁴/³ theta/cos²/³ theta = b² → (4)
Multiply (2) and (4), we get
(cos⁴/³ theta/sin²/³ theta)× (sin⁴/³ theta/cos²/³ theta) = a²b² → (5)
a² + b² =(cos⁴/³ theta/sin²/³ theta) + (sin⁴/³ theta/cos²/³ theta)
(cos² theta + sin² theta)/(sin²/³ theta cos²/³ theta)
= 1/sin²/³ theta cos²/³ theta
Consider, a²b²(a²+b²) = (sin²/³ theta cos²/³ theta) × 1/sin²/³ theta cos²/³ theta
= 1