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# Assignment # 01-- MTH101 (Spring 2014) Due Date# 27 -05-2014

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Our main purpose here discussion not just Solution

We are here with you hands in hands to facilitate your learning and do not appreciate the idea of copying or replicating solutions.

 Trigonometric Identities (Math | Trig | Identities)

 sin(theta) = a / c csc(theta) = 1 / sin(theta) = c / a cos(theta) = b / c sec(theta) = 1 / cos(theta) = c / b tan(theta) = sin(theta) / cos(theta) = a / b cot(theta) = 1/ tan(theta) = b / a

sin(-x) = -sin(x)
csc(-x) = -csc(x)
cos(-x) = cos(x)
sec(-x) = sec(x)
tan(-x) = -tan(x)
cot(-x) = -cot(x)

 sin^2(x) + cos^2(x) = 1 tan^2(x) + 1 = sec^2(x) cot^2(x) + 1 = csc^2(x) sin(x y) = sin x cos y cos x sin y cos(x y) = cos x cosy sin x sin y

tan(x y) = (tan x tan y) / (1  tan x tan y)

sin(2x) = 2 sin x cos x

cos(2x) = cos^2(x) - sin^2(x) = 2 cos^2(x) - 1 = 1 - 2 sin^2(x)

tan(2x) = 2 tan(x) / (1 - tan^2(x))

sin^2(x) = 1/2 - 1/2 cos(2x)

cos^2(x) = 1/2 + 1/2 cos(2x)

sin x - sin y = 2 sin( (x - y)/2 ) cos( (x + y)/2 )

cos x - cos y = -2 sin( (x - y)/2 ) sin( (x + y)/2 )

Trig Table of Common Angles
angle 0 30 45 60 90
sin^2(a) 0/4 1/4 2/4 3/4 4/4
cos^2(a) 4/4 3/4 2/4 1/4 0/4
tan^2(a) 0/4 1/3 2/2 3/1 4/0

Given Triangle abc, with angles A,B,C; a is opposite to A, b opposite B, c opposite C:

a/sin(A) = b/sin(B) = c/sin(C) (Law of Sines)

 c^2 = a^2 + b^2 - 2ab cos(C)b^2 = a^2 + c^2 - 2ac cos(B) a^2 = b^2 + c^2 - 2bc cos(A)
(Law of Cosines)

(a - b)/(a + b) = tan [(A-B)/2] / tan [(A+B)/2] (Law of Tangents)

cos(2x) = cos^2(x) - sin^2(x) = 2 cos^2(x) - 1 = 1 - 2 sin^2(x)

assignment questions

In other words we can “factor” a multiplicative constant out of a limit.

Question N0.1

Question No.2

part (1)

sec part of question  kese solve kia apne ?

Question No.3
the function is continuous because limit and function are same 0 value. limit approach to 0 (so no jump)

so, it is continuous

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