Question

look at the pic and give the answer

Answer 1

sorry mate i don't know the ans...

Answer 2

Step-by-step explanation:

If cosec theta - sin theta = m3 and sec theta - cos theta = n3 prove that

m^4n^2 + m^2n^4 = 1  

• Given cosec theta – sin theta = m^3
•         1/sin theta – sin theta = m^3
•        1 – sin^2 theta / sin theta = m^3
•       Cos^2 theta / sin theta = m^3 --------------- 1
•     (cos^2 theta / sin theta)^2/3 = (m^3)^2/3
•     So cos^4/3 theta / sin^2/3 theta = m^2 -----------2
• Let sec theta – cos theta = n^3
• 1/cos theta – cos theta = n^3
• 1 – cos^2 theta / cos theta = n^3
• So (sin^2 theta / cos theta) = n^3-----------3
• So (sin^2 theta / cos theta)^2/3 = (n^3)^2/3
• So sin^4/3 theta / cos^2/3 theta = n^2 ----------4
• Multiply 2 and 4 we get  
• (Cos^4/3 theta / sin^2/3 theta) x (sin^4/3 theta / cos^2/3 theta) = a^2b^2-----------5
• So a^2 + b^2 = (cos^4/3 theta / sin^2/3 theta) + (sin^4/3 theta / cos^2/3 theta)
•   (cos^2 theta + sin^2 theta) / (sin^2/3 theta cos^2/3 theta)  
• = 1/sin^2/3 theta cos^2/3 theta
• So m^2n^2 (m^2 + n^2) = 1
• Or m^4n^2 + m^2n^4 = 1

Reference link will be

https://brainly.in/question/682984