Question

if x sin³Θ+ycos³Θ=sinΘcosΘ and xsinΘ=ycosΘ, prove that x²+y²=1

Answer 1

✴✴ Hey friends!!✴✴

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✴✴ Here is your answer↓⬇⏬⤵

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▶⏩ It is given that:-)

=> xsinA = ycosA.

▶⏩To prove:-)

=> x² + y² = 1.

▶⏩ xsin³A + ycos³A = sinAcosA.

↪➡ (xsinA) sin²A + (ycosA) cos²A = sinAcosA.

↪➡ (xsinA) sin²A + (xsinA) cos²A = sinAcosA. [ ycosA = xsinA].

↪➡ (xsinA) (sin²A + cos²A) = sinAcosA.

↪➡ xsinA = sinAcosA...........(1).
[ Sin²A + Cos²A = 1 ].

↪➡ x = cosA
[ Squaring both side ].
↪➡ x² = cos²A ...................(2).

▶⏩ Now,

↪➡ xsinA = ycosA.

↪➡ cosAsinA = ycosA.......[ From equal(1)].

↪➡ y = sinA.
[ Squaring both side ].

↪➡ y² = sin²A ...................(3).

▶⏩ Add in equation (2) and (3).

↪➡ x² + y² = sin²A + cos²A.

↪➡ x² + y² = 1.✔✔

✴✴ Hence, it is proved ✅✅.

✴✴ Thanks!!✴✴.

☺☺☺ Hope it is helpful for you ✌✌✌.