1.

If `x=(sin^3P)/cos^2P,y=cos^3P/sin^2P" and " sinP+cosP=1/2` then find the value of x + y.

Answer» `x+y=(sin^5P+cos^5P)/(cos^2P sin^2P)`
Now, `sin^5P+cos^5P=(sinP+cosP)(sin^4P-sin^3PcosP+sin^2Pcos^2P-sinPcos^3P+cos^4P)`
`=(sinP+cosP)[(sin^4P+cos^4P)-sinPcosP(sin^2P+cos^2P)+sin^2Pcos^2P]`
`=(sinP+cosP)[1-2sin^2Pcos^2P-sinPcosP+sin^2Pcos^2P]`
`=(sinP+cosP)[1-sin^2Pcos^2P-sinPcosP]`
Now, `sinP+cosP=1/2`
`rArr1+2sinPcosP=1/4`
`rArr sinPcosP=-3/8`
Using these values, we get
`x+y=(1/2[1-9/64+3/8])/(9/64)`
`=79/18`


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