Question

if cosecΦ - sinΦ = a³ , secΦ - cosΦ = b³ ,

then prove that a²b² ( a² + b² ) = 1

please solve it.

Answer 1

$\huge\bold{Answer}$

$<b><u>$

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.

Thanks

Answer 2

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.

Thanks

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