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€c 0 — sin 0 = a’; sec 0 - cos B = B, prove thatL P’ + b)) =1. 38 e e |
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Answer» Consider cosec theta - sin theta = a³⇒ !/sin theta - sin theta = a³⇒ 1 - sin² theta/sin theta = a³cos² 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 thetacos²/³ thetaConsider, a²b²(a²+b²) =(sin²/³ theta cos²/³ theta)× 1/sin²/³ theta cos²/³ theta= 1 Hence proved. please like my answer if you find it useful |
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