If i=√-1 then
\[Arg\left[ \frac{(1+i)^{2025}}{1+i^{2022}} \right] =\]
If i=√-1 then
\[Arg\left[ \frac{(1+i)^{2025}}{1+i^{2022}} \right] =\]\(\frac{-π}{4}\)
\(\frac{π}{4}\)
\(\frac{3π}{4}\)
\(\frac{-3π}{4}\)
The Correct Option is A
Solution and Explanation
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Concepts Used:
Trigonometric Identities
Various trigonometric identities are as follows:
Even and Odd Functions
Cosecant and Secant are even functions, all the others are odd.
- sin (-A) = – sinA,
- cos (-A) = cos A,
- cosec (-A) = -cosec A,
- cot (-A) = -cot A,
- tan (-A) = – tan A,
- sec (-A) = sec A.
Pythagorean Identities
- sin2θ + cos2θ = 1
- 1 + tan2θ = sec2θ
- 1 + cot2θ = cosec2θ
Periodic Functions
- T-Ratios of (2π + x)
sin (2π + x) = sin x,
cos (2π + x) = cos x,
tan (2π + x) = tan x,
cosec (2π + x) = cosec x,
sec (2π + x) = sec x,
cot (2π+x)=cotx. - T-Ratios of (π -x)
sin (π–x) = sin x,
cos (π–x) = - cos x,
tan (π–x) = - tan x,
cosec (π–x) = cosec x,
sec (π–x) = - sec x,
cot (π–x) = - cot x. - T-Ratios of (π+ x)
sin (π+x) = - sin x,
cos (π+x) = - cos x,
tan (π+x) = tan x,
cosec (π+x) = - cosec x,
sec (π+x) = - sec x,
cot (π+x) = cot x. - T-Ratios of (2π – x)
sin (2π–x) = - sin x,
cos (2n–x) = cos x,
tan (2π–x) = - tan x,
cosec (2π–x) = - cosec x,
sec (2π–x) = sec x,
cot (2π-x) = - cot x
Sum and Difference Identities
- T-Ratios of (x + y)
sin (x+y) = sinx.cosy + cosx.sin y
cos (x+y) = cosx.cosy – sinx.siny - T-Ratios of (x – y)
sin (x–y) = sinx.cosy – cos.x.sin y
cos (x-y) = cosx.cosy + sinx.siny
Product of T-ratios
- 2sinx cosy = sin(x+y) + sin(x–y)
- 2cosx siny = sin(x+y) – sin(x–y)
- 2 cosx cosy = cos(x+y) + cos(x–y)
- 2sinx.siny = cos(x–y) – cos(x+y)
T-Ratios of (2x)
sin2x = 2sin x cos x
cos 2x = cos2x – sin2x
= 2cos2x – 1
= 1 – 2sin2x
T-Ratios of (3x)
sin 3x = 3sinx – 4sin3x
cos 3x = 4cos3x – 3cosx