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3 years ago

Prove that root3 (cosec (20))-sec (20)=4

2 answers:

7 0

Root 3 /sin 20 - 1/cos 20

= (root3 cos 20 - sin 20 )/ (sin 20 cos 20)
(multiplying and dividing by 2)
= 2* (root3/2 cos 20 -1/2 sin 20)/ (sin 20 cos 20)

now sin 20 cos 20 = (2 sin 20 cos 20)/2 = (sin 40)/2
and writing root3/2 = sin 60, 1/2 = cos 60

= 2* (sin 60 cos 20 - cos 60 sin 20)/ (sin 40)/2

= 4* (sin (60-20))/sin 40

= 4* (sin 40)/sin 40
=4

Ask if any doubt in any step :)

Mashcka [7]3 years ago

3 0

Hello,

1) Rappels:

all angles are in degree.

cos(a+b)=cos(a)cos(b)-sin(a)sin(b)
2sin(a)cos(a)=sin(2a)
sin(a)=cos(90-a)
cos(30)=√3/2
sin(30)=1/2
2)
√3 cos(20)-sin(20)=2(√3/2 cos(20)-1/2 sin(20)
=2*(cos(30) *sin(20)-sin(30) sin(20))=2 cos(30+20)=2 cos(50)

sin(20) cos(20)=1/2 cos(40)=1/2 cos(50)

√3 (cosec (20))-sec (20) = √3 /sin(20)-1/cos(20)
=[√3*cos(20)-sin(20)]/(sin(20)cos(20))
=2 cos(50)/ (1/2cos(50))=4

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Any 10th grader solve it for 50 points

kkurt [141]

Answer:

$\frac{a}{p}\times (q-r)+\frac{b}{q}\times (r-p)+\frac{c}{r}\times (p-q)\neq 0$ is proved for the sum of pth, qth and rth terms of an arithmetic progression are a, b,and c respectively.

Step-by-step explanation:

Given that the sum of pth, qth and rth terms of an arithmetic progression are a, b and c respectively.

First term of given arithmetic progression is A

and common difference is D

ie., $a_{1}=A$ and common difference=D

The nth term can be written as

$a_{n}=A+(n-1)D$

pth term of given arithmetic progression is a

$a_{p}=A+(p-1)D=a$

qth term of given arithmetic progression is b

$a_{q}=A+(q-1)D=b$ and

rth term of given arithmetic progression is c

$a_{r}=A+(r-1)D=c$

We have to prove that

$\frac{a}{p}\times (q-r)+\frac{b}{q}\times (r-p)+\frac{c}{r}\times (p-q)=0$

Now to prove LHS=RHS

Now take LHS

$\frac{a}{p}\times (q-r)+\frac{b}{q}\times (r-p)+\frac{c}{r}\times (p-q)$

$=\frac{A+(p-1)D}{p}\times (q-r)+\frac{A+(q-1)D}{q}\times (r-p)+\frac{A+(r-1)D}{r}\times (p-q)$

$=\frac{A+pD-D}{p}\times (q-r)+\frac{A+qD-D}{q}\times (r-p)+\frac{A+rD-D}{r}\times (p-q)$

$=\frac{Aq+pqD-Dq-Ar-prD+rD}{p}+\frac{Ar+rqD-Dr-Ap-pqD+pD}{q}+\frac{Ap+prD-Dp-Aq-qrD+qD}{r}$

$=\frac{[Aq+pqD-Dq-Ar-prD+rD]\times qr+[Ar+rqD-Dr-Ap-pqD+pD]\times pr+[Ap+prD-Dp-Aq-qrD+qD]\times pq}{pqr}$

$=\frac{Arq^{2}+pq^{2} rD-Dq^{2} r-Aqr^{2}-pqr^{2} D+qr^{2} D+Apr^{2}+pr^{2} qD-pDr^{2} -Ap^{2}r-p^{2} rqD+p^{2} rD+Ap^{2} q+p^{2} qrD-Dp^{2} q-Aq^{2} p-q^{2} prD+q^{2}pD}{pqr}$

$=\frac{Arq^{2}-Dq^{2}r-Aqr^{2}+qr^{2}D+Apr^{2}-pDr^{2}-Ap^{2}r+p^{2}rD+Ap^{2}q-Dp^{2}q-Aq^{2}p+q^{2}pD}{pqr}$

$=\frac{Arq^{2}-Dq^{2}r-Aqr^{2}+qr^{2}D+Apr^{2} -pDr^{2}-Ap^{2}r+p^{2}rD+Ap^{2}q-Dp^{2}q-Aq^{2}p+q^{2}pD}{pqr}$

$\neq 0$

ie., $RHS\neq 0$

Therefore $LHS\neq RHS$

ie., $\frac{a}{p}\times (q-r)+\frac{b}{q}\times (r-p)+\frac{c}{r}\times (p-q)\neq 0$

Hence proved

5 0

3 years ago

The manufacturing of a ball bearing is normally distributed with a mean diameter of 22 millimeters and a standard deviation of .

Misha Larkins [42]

Answer:

0.1507 or 15.07%.

Step-by-step explanation:

We have been given that the manufacturing of a ball bearing is normally distributed with a mean diameter of 22 millimeters and a standard deviation of .016 millimeters. To be acceptable the diameter needs to be between 21.97 and 22.03 millimeters.

First of all, we will find z-scores for data points using z-score formula.

$z=\frac{x-\mu}{\sigma}$ , where,