Question

cosec-sin=m and sec - cos= n show (m²n)⅔- (mn²)⅔ =1

Answer 1

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