Question

sin theta-cos theta = root 2 cos theta (only 2 is under root not cos theta) then diagnose sin theta +cos theta

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Answer 1

Answer:

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Answer 2

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$\bf\Huge\red{\mid{\overline{\underline{ ANSWER }}}\mid }$

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$\Large{\fbox{\color{purple}{QUESTION}}}$

sin θ - cos θ  = √2 cos θ

then, find sin θ + cos θ  = ?

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$\Large{\fbox{\color{purple}{ANSWER;-}}}$

√2 sin θ

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$\Large{\fbox{\color{Pink}{STEPS;-}}}$

--> sin θ - cos θ  = √2 cos θ

--> sin θ = √2 cos θ + cos θ

=> sin θ = ( √2 + 1 ) cos θ

=> [ sin θ / ( √2 + 1 ) ] = cos θ

0_0 --> We rationalized the

denominator in the 2nd step ^_^ 

[So, (a + b)(a - b) = (a^2 - b^2)]

=> [ sin θ ( √2 - 1 ) / ( 2 - 1 ) ] = cos θ

=> [ √2 sin θ - sin θ ] = cos θ

=> [ √2 sin θ - sin θ ] = cos θ

=> cos θ + sin θ = √2 sin θ

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