If xsin³A + ycos³A = sinAcosA and xsinA = ycosA , prove that
x² + y² = 1.
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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 ✌✌✌.
------------------------------------------------------
✴✴ Here is your answer↓⬇⏬⤵
⬇⬇⬇⬇⬇⬇⬇⬇⬇⬇⬇⬇⬇⬇⬇
⤵⤵⤵⤵⤵⤵⤵⤵⤵⤵⤵⤵⤵⤵⤵
▶⏩ 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 ✌✌✌.
Mayank765:
superb answer
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