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Questions  

The value of sin11213sin135 is equal to

a
π−sin−1⁡6365
b
π2−sin−1⁡5665
c
π2−cos−1⁡965
d
π−cos−1⁡3365

detailed solution

Correct option is B

Wehave, sin−1⁡1213−sin−1⁡35 =sin−1⁡12131−352−351−12132∵sin−1⁡x−sin−1⁡y=sin−1⁡x1−y2−y1−x2  if x2+y2≤1 orif xy>0 and x2+y2>1∀x,y∈[−1,1] =sin−1⁡1213×45−35×513=sin−1⁡48−1565=sin−1⁡3365=cos−1⁡1−33652=cos−1⁡31364225 ∵sin−1⁡x=cos−1⁡1−x2=cos−1⁡5665=π2−sin−1⁡5665∵sin−1⁡θ+cot−1⁡θ=π2

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