AP EAMCET Engineering 2013 Solved Paper

Given, sin‌θ+cos‌θ=p .......(i)
and sin3θ+cos3θ=q .......(ii)
⇒ (sin‌θ+cos‌θ) (sin2θ−sin‌θ.cos‌θ+cos2θ)=q
⇒ p(1−sin‌θ.cos‌θ)=q
[From Eq. (i) and sin2θ+cos2θ=1 ]
⇒ 1−sin‌θ.cos‌θ=

q

p

⇒ sin‌θ.cos‌θ=1−

q

p

......(iii)
On squaring both sides of Eq. (i), we get
sin2θ+cos2θ+2‌sin‌θ.cos‌θ=p2
⇒ 1+2(1−

q

p

)=p2 [from Eq. (iii)]
⇒ p+2(p−q)=p3
⇒ 3p−2q=p3
⇒ p3−3p=−2q
⇒ p(p2−3)=−2q