Question

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Oues : If a sinΦ + b cos Φ = c, then prove that a cos Φ - b sin Φ = ± √a² + b² -c². ​
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Answer 1

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Answer 2

Hi !

From given :- asin¢ + bcos¢ = c ____(1)

we have to prove acos¢ - bsin¢ = +-√a²+b²-c²

Let's assume , acos¢ - bsin¢ = x ____(2)

and , taking whole square of (1) and (2) and simply add them .

Then , (asin¢ + bcos¢ ) ² + (acos¢ - bsin¢ )² = c²+x²

=> a²sin²¢ + b²cos²¢ + 2absin¢*cos¢ + a²cos²¢ + b²sin²¢ - 2absin¢ cos¢ = x²+c²

=> a²sin²¢ + a²cos²¢ + b²sin²¢ + b²cos²¢ = x²+c²

Here , cancelled , 2absin¢ *cos¢

Now, => a²(sin²¢ + cos²¢ ) + b²(sin²¢ +cos²¢ ) = x²+c²

°•° sin²¢ + cos²¢ = 1

=> a²+b² = x²+c²

=> a²+b²-c² = x²

=> x = +-√a²+b²-c²

From (2) x = acos¢ - bsin¢

so, here proved , acos¢ - bsin¢ = +-√a²+b²-c²

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