Question

$\red{ \bold{hello}}$
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

given,

asin@+bcos@=c

• a²sin²@+b²cos²@+2absin@cos@=c²
• 2absin@cos@= c²-a²sin²@-b²cos²@-------1

then,

(acos@-bsin@)² = a²cos²@+b²sin²@ - 2absin@cos@

(acos@-bsin@)² = a²cos²@+ b²sin²@- c² +a²sin²@+b²cos²@

// putting the value from equation 1. (acos@-bsin@)²= a²(cos²@+sin²@) + b²(sin²@+cos²@) -c²

(acos@-bsin@)² = a² +b²-c²

then,

acos@-bsin@= (+_)(a²+b²-c²)^½

(proved).