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Harlamova29_29 [7]
3 years ago
12

Use trigonometric identities to simplify each expression. 1/cot^2 (x) - 1/cos^2(x)

Mathematics
1 answer:
PtichkaEL [24]3 years ago
7 0

Answer:

- 1

Step-by-step explanation:

\frac{1}{\cot ^2\left(x\right)}-\frac{1}{\cos ^2\left(x\right)}\\\\\mathrm{Use\:the\:basic\:trigonometric\:identity}:\quad \frac{1}{\cos \left(x\right)}=\sec \left(x\right),\\\frac{1}{\cot ^2\left(x\right)}-\sec ^2\left(x\right)\\\\\mathrm{Use\:the\:basic\:trigonometric\:identity}:\quad \frac{1}{\cot \left(x\right)}=\tan \left(x\right),\\\tan ^2\left(x\right)-\sec ^2\left(x\right)\\\\=> \frac{\sin ^2\left(x\right)}{\cos ^2\left(x\right)}-\frac{1}{\cos ^2\left(x\right)}\\

=> \frac{\sin ^2\left(x\right)-1}{\cos ^2\left(x\right)}\\\\=> -\frac{\cos ^2\left(x\right)}{\cos ^2\left(x\right)}\\=> - 1

Hope  that  helps!

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2x=24

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yawa3891 [41]

Answer:

1i) nth term = 12(-b/4)^(n-1)

ii) nth term = 3(-1/9)^(n-1)

2i) Sixth term= (-3b^5)/256

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Question:

The complete question as found on brainly (question-10746111):

Write down the nth term of each of the following G.Ps whose first two terms are given as follows. Also find the term stated besides each G.P.

i) 12,-3b,......sixth term

ii) 3,-1/3,..........;8th term?

Step-by-step explanation:

1) To solve terms involving Geometric progressions (GPs), we would first state the variables that are to be incorporated in the formula.

i) 12,-3b,......;sixth term

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r = common ratio = 2nd term /1st term

r = -3b/12 = -b/4

Then we would find the nth term

See attachment for details

ii) 3,-1/3,..........;8th term

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r = (-⅓)/3 = (-⅓)(⅓) = -1/9

Then we would find the nth term

See attachment for details

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See attachment for details

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