JEE Advanced | 2015Trigonometric EquationHard
Question
If α = 3 sin-1
and β = 3 cos-1
where the inverse trigonometric functions take only the principal values, then the correct option(s) is(are)
and β = 3 cos-1 where the inverse trigonometric functions take only the principal values, then the correct option(s) is(are)Options
A.cos β > 0
B.sin β > 0
C.cos (α + β) > 0
D.cos α < 0
Solution
α = 3 sin-1
& β = 3 cos-1
∵
> 
⇒ sin-1 
> sin-1
⇒ 3 sin-1
> 3sin-1
= 
∴ α >
∴ cos α < 0
Now , β = 3 cos-1
∵
< 
⇒ 3 cos-1 
> 3cos-1 
∴ β > π
∴ cosβ < 0 & sin β < 0
Now, α is slightly greater than
& β is slightly greater than π
∴ cos (α + β) > 0
& β = 3 cos-1∵
> ⇒ sin-1 > sin-1⇒ 3 sin-1
> 3sin-1 = ∴ α >
∴ cos α < 0
Now , β = 3 cos-1
∵
< ⇒ 3 cos-1 > 3cos-1 ∴ β > π
∴ cosβ < 0 & sin β < 0
Now, α is slightly greater than
& β is slightly greater than π∴ cos (α + β) > 0
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