Answer (Detailed Solution Below)

Option 3 : -45°
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General Knowledge for All Defence Exams (शूरवीर): Special Live Test
20 Qs. 20 Marks 16 Mins

Detailed Solution

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Concept:

Principal Values of Inverse Trigonometric Functions:

Function Domain Range of Principal Value
sin-1 x [-1, 1] [-π/2, π/2]
cos-1 x [-1, 1] [0, π]
csc-1 x R - (-1, 1) [-π/2, π/2] - {0}
sec-1 x R - (-1, 1) [0, π] - {π/2}
tan-1 x R (-π/2, π/2)
cot-1 x R (0, π)

 

Inverse Trigonometric Functions for Negative Arguments:

sin-1 (-x) - sin-1 x cos-1 (-x) π - cos-1 x
cosec-1 (-x) - cosec-1 x sec-1 (-x) π - sec-1 x
tan-1 (-x) - tan-1 x cot-1 (-x) π - cot-1 x

 

Calculation: 

As we know sin-1 (-x) = - sin-1 x

So,  

Let  = θ 

⇒ sin θ =  = sin 45° 

∴ θ = 45° 

Hence,  = -θ = -45°

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