Question

if x=sin^3p/cos^2p and y=cos^3p/sin2p and sinp +cosp =1/2 then value of x+y​

Answer 1

Given : x = Sin³p/Cos²p   , y = Cos³p/Sin²p , Sinp +Cosp =1/2

To find :   x + y

Solution:

x = Sin³p/Cos²p  

y = Cos³p/Sin²p

Sinp +Cosp =1/2

Squaring both sides

Sin²p +Cos²p  + 2SinpCosp= 1/4

=>1 + 2SinpCosp= 1/4

=> 2SinpCosp= -3/4

=> SinpCosp= -3/8

x + y  = Sin³p/Cos²p   + Cos³p/Sin²p

= (Sin³pSin²p  + Cos²pCos³p)/Cos²pSin²p

Sin³pSin²p  + Cos²pCos³p

= Sin³p(1 - Cos²p) + (1-Sin²p )Cos³p

=Sin³p - Sin³pCos²p + Cos³p - Sin²pCos³p

= Sin³p  + Cos³p  - Sin²pCos²p(Sinp + Cosp)

using a³ + b³ = (a + b)(a² + b² - ab)

= (Sinp + Cosp)(Sin²p + Cos²p - SinpCosp)  - (SinpCosp)²(Sinp + Cosp)

=  (Sinp + Cosp)(1 - SinpCosp -  (SinpCosp)²)

substitute SinpCosp= -3/8 and Sinp +Cosp =1/2

= (1/2)(1 - (-3/8) - (-3/8)²)

= (1/2)(1  + 3/8 - 9/64)

= (1/2)(64 + 24 - 9)/64

= 79/128

x + y  =  (Sin³pSin²p  + Cos²pCos³p)/Cos²pSin²p

= (79/128)/(-3/8)²

= 79/18

x + y  =  79/18

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