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saw5 [17]
3 years ago
12

1-sin^2x/sinx-cscx How do I solve this problem? I keep getting the wrong answer.

Mathematics
1 answer:
Annette [7]3 years ago
3 0

Answer:

-sinx

Step-by-step explanation:

a trig identity that is crucial to solving this problem is: sin^2 + cos^2 = 1

with knowing that, you can manipulate that and turn it into 1 - sin^2x = cos^x

so 1-sin^2x/sinx - cscx becomes cos^2x/sinx - cscx

it is also important to know that cscx is the same thing as 1/sinx

knowing this information, cscx can be replaced with 1/sinx

(cos^2x)/(sinx - 1/sinx)

now sinx and 1/sinx do not have the same denominator, so we need to multiply top and bottom of sinx by sinx; it becomes....

cos^2x

---------------------

(sin^2x - 1)/sinx

notice how in the denominator it has sin^2x-1 which is equal to -cos^2x

so now it becomes:

cos^2x

--------------

-cos^2x/sinx

because we have a fraction over a fraction, we need to flip it

cos^2x          sinx

---------- * ----------------

1                  -  cos^2x

because the cos^2x can cancel out, it becomes 1

now the answer is -sinx

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The heights of young women aged 20 to 29 follow approximately the N(64, 2.7) distribution. Young men the same age have heights d
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I.e. ф-1 [(M-64)/2.7] = 0.75 i.e. (M-64)/2.7 = ф-1 (0.75) = 0.675 i.e. M = 65.8198 inches

Let R be the random variable denoting the height of young men

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i.e. (R-69.3)/2.8 – N(0,1)

therefore the probability required = P(R ˂65.8198) = P[(R-69.3)/2.8 ˂ (65.8198 – 69.3)/2.8]

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