Question

HEYA GUYS
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PLEASE SOLVE IT
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If cosec theta -- sin theta = m, and
sec theta -- cos theta = n,
Prove that (m^2n)^2/3 + (mn^2)^2/3 = 1​

Answer 1

Answer:

Step-by-step explanation:

Solution:-

Given cosec theta - sin theta = m, sec theta - cos theta = n

Given that cosec theta - sin theta = m

→ !/sin theta - sin theta = m

⇒ (1-sin² theta)/sin theta = m → cos² theta/sin theta = m

and sec theta - cos theta = n

⇒ 1/cos theta - cos theta = n → (1-cos² theta)/cos theta = n

sin² theta/cos theta = n

Now (m²n)²/³ + (mn²)²/³

⇒ (cos⁴ theta/sin² theta × sin² theta/cos theta)²/³ + (cos² theta/sin theta × sin⁴ theta/cos² theta)²/³

⇒ (cos³ theta)²/³ + (sin³ theta)²/³

⇒cos² theta + sin² theta

= 1 Hence proved