anonymous one year ago Can someone check my answer, I feel like I did it completely wrong!! 13. Verify the identity. Justify each step. sec (0) sec(0) ------------- - ------------- = 2csc * (0) csc(0)-cot(0) csc(0)+cot(0) (sec(0)*(csc(0)+ cot(0)) - (sec(0)*(csc(0)-cot(0))) --------------------------------------------------- = 2csc*(0) (csc(0)-cot(0))*(csc(0)*(csc(0)+cot(0)) (sec(0)*csc(0)+cot(0)*sec(0))-(2sec(0)*csc(0)-cot(0)*sec(0)) ------------------------------------------------------------ (csc^2(0)-cot^2(0)) =2csc*(0) (2csc*(2theta)+csc(0))-(2csc*(2theta)-csc(0))

1. anonymous

2. anonymous

sec (0) sec(0) ------------- - ------------- = 2csc * (0) csc(0)-cot(0) csc(0)+cot(0)

3. anonymous

(sec(0)*(csc(0)+ cot(0)) - (sec(0)*(csc(0)-cot(0))) --------------------------------------------------- = 2csc*(0) (csc(0)-cot(0))*(csc(0)*(csc(0)+cot(0))

4. anonymous

(sec(0)*csc(0)+cot(0)*sec(0))-(2sec(0)*csc(0)-cot(0)*sec(0)) ------------------------------------------------------------ (csc^2(0)-cot^2(0)) =2csc*(0)

5. anonymous

(2csc*(2theta)+csc(0))-(2csc*(2theta)-csc(0)) --------------------------------------------- = 2csc*(0) (csc^2(0)-cot^2(0))

6. anonymous

csc^2(0)-cot^2(0)=1 ((2csc*(2theta)+csc(0))-2csc*(2theta)+csc(0)) 2csc(0)=2csc(0)

7. Nnesha

confusion and confusion o,o how did you get 2 there http://prntscr.com/8qb8cf ?

8. anonymous

im not quite sure. I watched a video about how to do it but i feel like i didnt learn a thing lol could you help me

9. Nnesha

hmm sure i'll just start from http://prntscr.com/8qb8um not gonna start over by different way

10. anonymous

omg thank you! you are a lifesaver. my teacher already thinks im crazy with some of the answers i give her

11. Nnesha

distribute $\large\rm \frac{ \sec* \csc + \sec*\cot -\sec*\csc + \sec *\cot }{ \csc^2 \theta - \cot^2 \theta }$ ignore theta for a sec

12. Nnesha

what's the definition of sec? like csc = 1/sin so sec = ?

13. anonymous

im not sure:/

14. Nnesha

oh okay thats fine but these are very important $\sin \theta= \frac{1}{\csc \theta} ~~~\cos \theta =\frac{1}{\sec \theta }~~~ \tan \theta= \frac{1}{\cot \theta }$ sin cos tan and csc sec cot are reciprocal

15. Nnesha

$\large\rm \frac{ \sec* \csc + \sec*\cot -\sec*\csc + \sec *\cot }{ \csc^2 \theta - \cot^2 \theta }$ first look at the numerator combine like terms !let me know what u get :=)

16. Nnesha

like sec*csc and -sec*csc are like terms so sec*csc - sec*csc = ?

17. anonymous

18. Nnesha

that is a different question right ?

19. anonymous

i redid the equation

20. Nnesha

i'm confused don't know how u got tan and cos at the beginning

21. Nnesha

the work u posted here is almost correct but there was a little mistake so i'll say keep the same method

22. Nnesha

$\large\rm \frac{ \sec* \csc + \sec*\cot -\sec*\csc + \sec *\cot }{ \csc^2 \theta - \cot^2 \theta }$ just two steps then we are done !

23. anonymous

oh shoot it was the wrong one im sorry. i guess ive been doing too much work today apparently lol ill just start from where you lfet off then

24. Nnesha

ohh i was so confused.

25. anonymous

26. Nnesha

its okay $\large\rm \frac{ \sec* \csc + \sec*\cot -\sec*\csc + \sec *\cot }{ \csc^2 \theta - \cot^2 \theta }$ can you simplify the numerator just like simple algebra :=) combine like terms :=)

27. anonymous

wouldnt it cancel eachother out because sec*csc+sec*cot is repeated twice so it would be sec*csc+sec*cot as the numerator

28. Nnesha

yes right sec*cot + sec *cot = 2 sec*cot $\huge\rm \frac{ 2\sec*\cot }{ \sec^2 - \cot^2}$ we can convert sec to 1/cos and cot to cos/sin

29. Nnesha

cot = cos /sin and tan = sin/cos bot hare reciprocal of each other

30. Nnesha

oh well latex doesn't work |dw:1444619074859:dw|

31. Nnesha

use the identity cot^2 (x) +1 = csc^2 move the cot^2 to the right side what would you get ?

32. Nnesha

$\huge\rm \frac{ 2(\frac{1}{\cos}*\frac{\cos}{\sin}) }{ \sec^2 \theta -\cot^2 }$ simplify the numerator if we solve cot^2 theta +1 =csc^2 for 1 we would get csc^2 -cot^2 =1 so you can replace csc^2 -cot^2 with 1 simplify the numerator

33. anonymous

okay

34. Nnesha

http://math2.org/math/trig/identities.htm copy these down on ur notes identities gt g

35. anonymous

would i distribute. like 2 x the things in the parentheses

36. anonymous

@jim_thompson5910 could you help with the last bit?

37. anonymous

@shaik0124

38. anonymous

what are we trying to solve. I'm confused.

39. anonymous

is it the picture you posed?

40. anonymous

posted*

41. anonymous

yes the very first picture posted. I have to verify the identity and justify each step

42. anonymous

oh right

43. anonymous

i was told that i had it mostly right. we were working from the second line or the third post but i think she started having connection issues

44. anonymous

let me just do it by hand from the beginning. one sec :)

45. anonymous

okay thank you so much! this is the last problem that I have to fix and weve been working on it for probably an hour

46. anonymous

hahaa! i'm sure we will get it done!

47. anonymous

thank you lol im going on about 18 hours doing school work today!! -_-

48. anonymous

ok sweet, i got it! lets do this

49. anonymous

alright:)

50. anonymous

when i approach these question, what i tend to do is solve the massive weird stuff and compare it to the side which is simplified. so in other words, what i would do with this question is to simplify the left hand side and compare it to the right hand side, right?

51. anonymous

yes makes sense

52. anonymous

good! so lefts ONLY work with the LHS (left hand side)

53. anonymous

$\frac{ \sec(\theta }{ \csc(\theta)-\cot(\theta) }-\frac{ \sec(\theta) }{ \csc(\theta)+\cot(\theta) }$

54. anonymous

which would cancel each other out right

55. anonymous

now, my approach would be to get the denominators the same.

56. anonymous

see how our denominators are a bit different

57. anonymous

oh yea ones (-) and the others (+)

58. anonymous

we can only subtract fractions if there denominators are the same

59. anonymous

so we are going to multiply the first fraction by the second denominator and vide versa...watch $\frac{ \sec(\theta)\left[ \csc(\theta)+\cot(\theta) \right] }{(\csc(\theta) -\cot(\theta))\left[ \csc(\theta)+\cot(\theta) \right] }- \frac{ \sec(\theta)\left[ \csc(\theta)-\cot(\theta) \right] }{ \left[ \csc(\theta)+\cot(\theta) \right] \left[ \csc(\theta)-\cot(\theta) \right]}$

60. anonymous

sorry i took too long with syntax

61. anonymous

thants fine im willing to wait to learn

62. anonymous

pretty much what this is: $\frac{ A }{ B }+\frac{ C }{ D }=\frac{ AD }{ BD }+\frac{ CB }{ DB }=\frac{ AD+CB }{ BD }$

63. anonymous

right, i just want to make sure you are on the right track

64. anonymous

got it?

65. anonymous

yes im doing it right now

66. anonymous

now remember the technique of difference of two squares rule: $(X+Y)(X-Y)=X^2-Y^2$

67. anonymous

so this means, $\left[ \csc(\theta)-\cot(\theta) \right]\left[ \csc(\theta)+\cot(\theta) \right]=\csc ^{2}(\theta)-\cot ^{2}(\theta)$

68. anonymous

right?

69. anonymous

right I was just typing it in now lol it takes me a little longer to write it out on paper

70. anonymous

and thats our denominator right

71. anonymous

yep! thats our denominator, good!

72. anonymous

so lets simplify the big equation that i had posted

73. anonymous

which one?

74. anonymous

$\frac{ \sec(\theta)\left[ \csc(\theta)+\cot(\theta) \right] }{(\csc(\theta) -\cot(\theta))\left[ \csc(\theta)+\cot(\theta) \right] }- \frac{ \sec(\theta)\left[ \csc(\theta)-\cot(\theta) \right] }{ \left[ \csc(\theta)+\cot(\theta) \right] \left[ \csc(\theta)-\cot(\theta) \right]}$

75. anonymous

that one

76. anonymous

okay that looks better lol

77. anonymous

omg my syntax keeps screwing up

78. anonymous

just had to refresh my page haha a

79. anonymous

$\frac{ \sec(\theta)\left[ \csc(\theta)+\cot(\theta) \right] }{ \csc ^{2}(\theta)-\cot ^{2}(\theta) }-\frac{ \sec(\theta) \left[ \csc(\theta)-\cot(\theta) \right]}{ \csc ^{2}(\theta)-\cot ^{2}(\theta) }$

80. anonymous

good?

81. anonymous

so lets group everything now

82. anonymous

$\frac{ \sec(\theta)\left[ \csc(\theta)+\cot(\theta) \right] -\sec(\theta)\left[ \csc(\theta)-\cot(\theta) \right]}{ \csc ^{2}(\theta)-\cot ^{2}(\theta)}$

83. anonymous

expand the numerator

84. anonymous

$\frac{ \sec(\theta)\csc(\theta)+\sec(\theta)\cot(\theta)-\sec(\theta)\csc(\theta)+\sec(\theta)\cot(\theta) }{ \csc ^{2}(\theta)-\cot ^{2}(\theta) }$

85. anonymous

now simplify! the numerator

86. anonymous

okay

87. anonymous

$\frac{ 2\sec(\theta)\cot(\theta) }{ \csc ^{2}(\theta) -\cot ^{2}(\theta)}$ this is because sec(theta)csc(theta)-sec(theta)csc(theta)=0

88. anonymous

so we are left with this

89. anonymous

so t hey each cancel out leaving us with your fraction above?

90. anonymous

they*

91. anonymous

yep!

92. anonymous

and thats all that I need to do for the question? It doesnt need to be broken down anymore?

93. anonymous

nope we keep going

94. anonymous

okay i thought so

95. anonymous

did nnesha explain to you why $\csc ^{2}(\theta) -\cot ^{2}(\theta)=1$

96. anonymous

if so i dont quite understand still

97. anonymous

do you have a list of trig identities?

98. anonymous

yes

99. anonymous

http://www.purplemath.com/modules/idents.htm where it says $1 + \cot ^{2}(t) = \csc ^{2}(t)$

100. anonymous

so we just re-arrange that equation $\csc ^{2}(t) - \cot ^{2}(t)=1$

101. anonymous

so cot^2(t) is always going to = 1

102. anonymous

no, $\csc ^{2}(t) - \cot ^{2}(t)=1$

103. anonymous

oh i see what you mean now

104. anonymous

so where we how we have our denominator as $\csc ^{2}(t) - \cot ^{2}(t)$ we can just substitute that as 1 since we know the trig identity that $\csc ^{2}(t) - \cot ^{2}(t)=1$

105. anonymous

right?

106. anonymous

correct

107. anonymous

awesome :)

108. anonymous

so that means $\frac{ 2\sec(\theta)\cot(\theta) }{ \csc ^{2}(\theta) -\cot ^{2}(\theta)}$ REDUCES to: $\frac{ 2\sec(\theta)\cot(\theta) }{ 1}$ which is simply $2\sec(\theta)\cot(\theta)$

109. anonymous

understand that?

110. anonymous

yes because csc^2 (0)-cot^2(0)=1

111. anonymous

now all we need to know is some basic trig formulas: $\csc(\theta)=\frac{ 1 }{ \sin (\theta) }$ $\sec(\theta)=\frac{ 1 }{ \cos(\theta) }$ and $\cot(\theta)=\frac{ 1 }{ \tan(\theta) }=\frac{ \cos(\theta) }{ \sin(\theta) }$

112. anonymous

so $2\sec(\theta)\cot(\theta)=2\left( \frac{ 1 }{ \cos(\theta) } \right)\left( \frac{ \cos(\theta) }{ \sin(\theta) } \right)$ the cos(X) part cancels since we are multiplying fractions, so we are left with: $2\sec(\theta)\cot(\theta)=2\left( \frac{ 1 }{ \sin(\theta) } \right)$

113. anonymous

now we know that: $\csc(\theta)=\frac{ 1 }{ \sin(\theta)}$

114. anonymous

therefore $2\sec(\theta)\cot(\theta)=2\left( \frac{ 1 }{ \sin(\theta) } \right)=2\csc(\theta)$

115. anonymous

now we can see that the LHS reduces to $2\csc(\theta)$ and that means this is also the RHS!!!!!!

116. anonymous

which 2 ces(0) is the final answer

117. anonymous

YEP! so that means we are dun!

118. anonymous

So for what the teachers asking i simply put everything we have figured out together for her answer

119. anonymous

so what we did was reduce the left hand side right? and we found that the left hand side equalled to the right hand side

120. anonymous

yes that was the first step we did

121. anonymous

Working with the LHS, we want to reduce this to see if it equals the RHS. LHS IMPLIES: $\frac{ \sec(\theta }{ \csc(\theta)-\cot(\theta) }-\frac{ \sec(\theta) }{ \csc(\theta)+\cot(\theta) }$ Getting denominators the same: $\frac{ \sec(\theta)\left[ \csc(\theta)+\cot(\theta) \right] }{(\csc(\theta) -\cot(\theta))\left[ \csc(\theta)+\cot(\theta) \right] }- \frac{ \sec(\theta)\left[ \csc(\theta)-\cot(\theta) \right] }{ \left[ \csc(\theta)+\cot(\theta) \right] \left[ \csc(\theta)-\cot(\theta) \right]}$ Now from denominators, we can use the difference of two squares: $\frac{ \sec(\theta)\left[ \csc(\theta)+\cot(\theta) \right] }{ \csc ^{2}(\theta)-\cot ^{2}(\theta) }-\frac{ \sec(\theta) \left[ \csc(\theta)-\cot(\theta) \right]}{ \csc ^{2}(\theta)-\cot ^{2}(\theta) }$ Now subtract the two fractions since we have the denominators the same: I.e get it into one fraction. $\frac{ \sec(\theta)\csc(\theta)+\sec(\theta)\cot(\theta)-\sec(\theta)\csc(\theta)+\sec(\theta)\cot(\theta) }{ \csc ^{2}(\theta)-\cot ^{2}(\theta) }$ Simplifying the numerator we get: $\frac{ 2\sec(\theta)\cot(\theta) }{ \csc ^{2}(\theta) -\cot ^{2}(\theta)}$ Now, using trig identities for the numerator where $\csc ^{2}(\theta) -\cot ^{2}(\theta)=1$ Therefore, it reduces to $\frac{ 2\sec(\theta)\cot(\theta) }{ 1}$ Using basic trig identities, we can simplify in terms of sin(x) and cos(x) $2\sec(\theta)\cot(\theta)=2\left( \frac{ 1 }{ \cos(\theta) } \right)\left( \frac{ \cos(\theta) }{ \sin(\theta) } \right)$ Cancelling out the cos(theta) we get, $2\sec(\theta)\cot(\theta)=2\left( \frac{ 1 }{ \sin(\theta) } \right)$ finally we know $\csc(\theta)=\frac{ 1 }{ \sin(\theta)}$ therefore $2\sec(\theta)\cot(\theta)=2\left( \frac{ 1 }{ \sin(\theta) } \right)=2\csc(\theta)$ =RHS (As required)

122. anonymous

:)

123. anonymous

THANK YOU SO MUCH!!! You are a huge lifesaver!!! This was my last question I had to correct and understand for me to complete my diploma. You truly deserve best response for your very detailed help! Thank you thank you!!!

124. anonymous

: ) you're too kind

125. anonymous

I gave you best response and a raving testimonial on your page!

126. anonymous

hahahaha if you want you can fan me so if you stuck with trig equations, you can easily find me if i'm online :) well done though. they are a bit tricky, but there is always a systematic approach!

127. anonymous

Im already a fan(; If i have any other questions I will be sure to ask you! thank you again and have a wonderful night/morning (its 1am here)

128. anonymous

jeez. i live in australia so its 3.30pm here haha

129. anonymous

im in michigan,usa lol

130. anonymous

i should be doing lecture writing, but i procrastinate on open study

131. anonymous

ahah jeez. thats hectic

132. anonymous

see im good at writting so if you need help with THAT let me know lol I will be on tomorrow but right now..im off to bed! my two year old wakes up too early lol

133. anonymous

haha good night

134. anonymous

goodnight(;

135. Nnesha

$$\color{blue}{\text{Originally Posted by}}$$ @Nnesha $\huge\rm \frac{ 2(\frac{1}{\cos}*\frac{\cos}{\sin}) }{ \sec^2 \theta -\cot^2 }$ simplify the numerator if we solve cot^2 theta +1 =csc^2 for 1 we would get csc^2 -cot^2 =1 so you can replace csc^2 -cot^2 with 1 simplify the numerator $$\color{blue}{\text{End of Quote}}$$ we replace sec with 1/sec and cot with cos/sin se cancel out the cso theta $\huge\rm \frac{ 2(\frac{1}{\cancel{cos}}*\frac{\cancel{cos}}{\sin}) }{ \sec^2 \theta -\cot^2 }$ and csc =1/sin like i said they are reciprocal as mentioned above sec^2 +1 = cot^2 set it equal to one so we can replace sec^2 -cot^2 with 1

136. Nnesha

and yea sorry *net issues* :/