Question

If cosecx -sinx=m and secx-cosx=n , prove that (m^2)^2/3+(mn^2)^2/3=1

Answer 1

Actually, the LHS should be (m^2n)^(2/3)  + (mn^2)^(2/3).

Given, cosec x - sin x = m

            1/sin x - sin x = m

             (1-sin^2 x)/sin x = m

             cos^2 x/sinx = m  ------------ (1)


Given that sec x - cos x = n

                   1/cos x - cos x = n

                   1-cos^2 x/cos x = n

                    sin^2 x/cos x = n    ------------------ (2)


(m^2n)^2/3 + (mn^2)^2/3 

= (cos^4 x/sin^2 x  * sin^2 x/cos x)^2/3 + (cos^2 x/sin x * sin^4 x /cos^2 x)^2/3

= (cos^3 x)^2/3  + (sin^3 x)^2/3

= cos^2 x + sin ^2 x

= 1.


LHS = RHS.


Hope this helps!

Answer 2

Answer:

Given, cosec x - sin x = m

1/sin x - sin x = m

(1-sin^2 x)/sin x = m

cos^2 x/sinx = m ------------ (1)

Given that sec x - cos x = n

1/cos x - cos x = n

1-cos^2 x/cos x = n

sin^2 x/cos x = n ------------------ (2)

(m^2n)^2/3 + (mn^2)^2/3

= (cos^4 x/sin^2 x * sin^2 x/cos x)^2/3 + (cos^2 x/sin x * sin^4 x /cos^2 x)^2/3

= (cos^3 x)^2/3 + (sin^3 x)^2/3

= cos^2 x + sin ^2 x

= 1.