Answer (Detailed Solution Below)
Option 2 :
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Detailed Solution
Download Solution PDFव्याख्या:
यदि sinθ = x ⇒ θ = sin-1x, θ ∈ [-π/2, π/2] के लिए,
sin (sin-1 x) =x, -π/2 ≤ x ≤ π/2 के लिए
हमारे पास है,
= sin-1(sin(
∴ sin-1(sin(
Additional Informationव्युत्क्रम त्रिकोणमितीय फलनों का मुख्य मान:
फलन |
प्रांत |
मुख्य मान की परास |
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) |
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