Question
sin−1|cosx|−cos−1|sinx|=a has at least one solution if a ∈
Answer: Option A
:
A
sin−1|cosx|−cos−1|sinx|=π2−cos−1|cosx|−π2+sin−1|sinx|=a⇒sin−1|sinx|−cos−1|cosx|=a⇒a=0∀x
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:
A
sin−1|cosx|−cos−1|sinx|=π2−cos−1|cosx|−π2+sin−1|sinx|=a⇒sin−1|sinx|−cos−1|cosx|=a⇒a=0∀x
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