Question

Sin(22-theta)-cos(68+theta)is

Answer 1

Answer:

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Answer 2

Answer:

$Value \: of \: sin(22-\theta) - cos(68+\theta)=0$

Step-by-step explanation:

$Value \: of \: sin(22-\theta) - cos(68+\theta)\\=sin(22-\theta) - cos[(90-22)+\theta]$

$= sin(22-\theta) - cos[90 -(22-\theta)]$

$= sin(22-\theta) - sin(22-\theta)$

$Since, \boxed {cos(90-\theta)= sin\theta}$

$= 0$

Therefore,

$Value \: of \: sin(22-\theta) - cos(68+\theta)=0$

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