Question

please anwer my question​

Answer 1

$<font color=red><marquee behavior=alternate>Solution</marquee></font>$

a cos∅-b sin∅=c

{squaring on both sides}

=>( a cos∅-b sin∅)²=c²

=>a²cos²∅+b²sin²∅-2absin∅cos∅=c².....,...(i)

now....

a sin∅+b cos∅=y (let)

{squaring on both sides}

=>(a sin∅+b cos∅)²=y²

=>a²sin²∅+b²cos²∅+2ab sin∅ cos∅=y²......(ii) adding equations (i)&(ii)...we get,,,

a²cos²∅+b²sin²∅-2absin∅cos∅+a²sin²∅+b²cos²∅+2ab sin∅ cos∅=c²+y²

=>a²(sin²∅+cos∅)+b²(sin∅+cos∅)=c²+y²

=>a²+b²-c²=y²

$= > y = \sqrt{a {}^{2} + b {}^{2} - c {}^{2} }$

=>a sin∅+b cos∅$= \sqrt{a {}^{2} + b {}^{2} - c {}^{2} }$ (proved)

Answer 2

Step-by-step explanation:

a cos∅-b sin∅=c

{squaring on both sides}

=>( a cos∅-b sin∅)²=c²

=>a²cos²∅+b²sin²∅-2absin∅cos∅=c².....,...(i)

now....

a sin∅+b cos∅=y (let)

{squaring on both sides}

=>(a sin∅+b cos∅)²=y²

=>a²sin²∅+b²cos²∅+2ab sin∅ cos∅=y²......(ii) adding equations (i)&(ii)...we get,,,

a²cos²∅+b²sin²∅-2absin∅cos∅+a²sin²∅+b²cos²∅+2ab sin∅ cos∅=c²+y²

=>a²(sin²∅+cos∅)+b²(sin∅+cos∅)=c²+y²

=>a²+b²-c²=y²

\begin{lgathered}= > y = \sqrt{a {}^{2} + b {}^{2} - c {}^{2} } \\\end{lgathered}

=>a sin∅+b cos∅

\begin{lgathered}= \sqrt{a {}^{2} + b {}^{2} - c {}^{2} } \\\end{lgathered}

(proved)