If cosec theta minus sin theta is equals to M cube and secant theta minus cos theta = 2 and cube then prove that and raise to power 4 and square + sin square x by 4 is equals to 1
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Answer:
HERE IS YOUR ANSWER
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.
CORRECT QUESTION :
If cosec theta - sin theta=m³ and sec theta - cos theta=n³, then prove that (m^2n)^2/3 + (mn^2)^2/3 = 1
cosec Ф - sin Ф = m³ .
⇒ 1/sin Ф - sin Ф = m³
⇒ ( 1 - sin²Ф )/sin Ф = m³
⇒ cos²Ф/sin Ф = m³
⇒ cot Ф cos Ф = m³
⇒ cos²Ф / sin Ф = m³ .........( 1 )
sec Ф - cos Ф = n³
⇒ 1/cosФ - cosФ = n³
⇒ ( 1 - cos²Ф )/cosФ = n³
⇒ sin²Ф/cosФ = n³ ...........( 2 )
We have to prove that :
(m²n)^2/3 + (mn²)^2/3 = 1
⇒ ( cos⁴Ф/sin²Ф . sin²Ф/cosФ )^(2/3) + ( sin⁴Ф/cos²Ф . cos²Ф/sinФ)^(2/3)
= cos²Ф + sin²Ф
= 1