Question

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Answer 1

Step-by-step explanation:

$\huge\underline{\underline{\ Question}}$

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$3 \sin( \theta) + 5 \cos( \theta) = 5 \\ prove \: that = > \\ 5 \sin( \theta) - 3 \cos( \theta) = |3|$

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$\huge\underline{\underline{\ Answer}}$

Let....

$5 \sin( \theta) - 3 \cos( \theta) = |a|$

$3 \sin( \theta) + 5 \cos( \theta) = 5$

$squaring \: both \: = >$

$25{ \sin}^{2} ( \theta) + 9 { \cos}^{2} ( \theta) - 30 \sin( \theta) \cos( \theta) = {a}^{2} \\ \\ 9 { \sin}^{2} ( \theta) + 25 { \cos}^{2} ( \theta) + 30 \sin( \theta) \cos( \theta) = 25 \\ Adding both =>$

$25{ \sin}^{2} ( \theta) + 9 { \cos}^{2} ( \theta) \cancel{- 30 \sin( \theta) \cos( \theta) }= {a}^{2} \\ \\ 9 { \sin}^{2} ( \theta) + 25 { \cos}^{2} ( \theta) \cancel { + 30 \sin( \theta) \cos( \theta) } = 25$

Also,

$\huge\boxed{{ \sin}^{2} ( \theta) + { \cos}^{2} ( \theta) = 1}$

$\implies 34({ \sin}^{2} ( \theta) + { \cos}^{2} ( \theta) )= 25 + {a}^{2} \\ \implies {a}^{2} = 9 \\ \implies a = |3| \\ \\ \huge\boxed{ 5 \sin( \theta) - 3 \cos( \theta) = |3| } \\ \\ hence \: proved.....$