If cosec theta - sin theta = a3 and sec theta - cos theta = b3 prove that a2b2 ( a2+ b2) = 1
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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 Hence proved.
⇒ !/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 Hence proved.
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