Question

Solve the ques 14 in the attachment

Standard:- 10

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Answer 1

Given,
cosecø -sinø = m
1/sinø - sinø = m
(1-sin²ø)/sinø = m
cos²ø/sinø= m
similarly
sin²ø/cosø= n
putting these values in LHS
refer to attachment.......

Answer 2

taking ∅ = x

Given, cosec x - sin x = m

1/sinx - sinx = m

(1-sin²x)/sin x = m

 cos²x/sinx = m  ------------ (1)

sec x - cos x = n

1/cos x - cos x = n

1 - cos²x/cos x = n

 sin²x/cos x = n    ------------------ (2)

now,

(m²n)⅔ + (mn²)⅔ 

(cos⁴x/sin²x  * sin²x/cos x)⅔ + (cos²x/sinx * sin⁴x /cos²x)⅔

(cos³x)⅔  + (sin³x)⅔

cos²x + sin²x

1