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3 years ago

If tan tetha =8/15,find the value of sin tetha+cos tetha all divided by cos tetha (1-cos tetha)

Mathematics

1 answer:

Dahasolnce [82]3 years ago

7 0

Answer: 195.5 or 12 7/32

Step-by-step explanation:

There is no letter tetha in the table so I use α instead. However it is not sence to final result.

The expression is:

(sinα+cosα)/(cosα*(1-cosα))

Lets divide the nominator and denominator by cosα

(sinα/cosα+cosα/cosα)/(cosα*(1-cosα)/cosα)= (tanα+1)/(1-cosα)=

=(8/15+1)/(1-cosα)= 23/(15*(1-cosα)) (1)

As known cos²α=1-sin²α (divide by cos²α both sides of equation)

cos²a/cos²α=1/cos²α-sin²α/cos²α

1=1/cos²α-tg²α

1/cos²α=1+tg²α

cos²α=1/(1+tg²α)

cosα=sqrt(1/(1+tg²α))= +-sqrt(1/(1+64/225))=+-sqrt(225/(225+64))=

=+-sqrt(225/289)=+-15/17 (2)

Substitute in (1) cosα by (2):

1st use cosα=15/17

1) 23/(15*(1-cosα)) =23/(15*(1-15/17))= 23*17/2=195.5

2-nd use cosα=-15/17

2)23/(15*(1-cosα)) =23/(15*(1+15/17))= 23*17/32=12 7/32

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Answer:

$T(t) = ( -sin(t^2), cos(t^2) )\\\\N(t) = T'(t) / |T'(t) | = (-cos(t^2) , -sin(t^2))$

$T(t) =r'(t)/|r'(t)| = (2t/ \sqrt{4t^2 + 36} , -6/\sqrt{4t^2 + 36},0)$

$N(t) =T(t)/|T'(t)| = (3/(9 + t^2)^{1/2} , t/(9 + t^2)^{1/2},0)$

Step-by-step explanation:

Remember that for any curve $r(t)$

The tangent vector is given by

$T(t) = \frac{r'(t) }{| r'(t)| }$

And the normal vector is given by

$N(t) = \frac{T'(t)}{|T'(t)|}$

For this case, using the chain rule

$r'(t) = ( -10*2tsin(t^2) , 102t cos(t^2) )\\$

And also remember that

$|r'(t)| = \sqrt{(-10*2tsin(t^2))^2 + ( 10*2t cos(t^2) )^2} \\\\ = \sqrt{400 t^2*( sin(t^2)^2 + cos(t^2) ^2 })\\=\sqrt{400t^2} = 20t$

Therefore

$T(t) = r'(t) / |r'(t) | = ( -10*2tsin(t^2) , 10*2t cos(t^2) )/ 20t\\\\ = ( -10*2tsin(t^2)/ 20t , 10*2t cos(t^2) / 20t )\\= ( -sin(t^2), cos(t^2) )$

Similarly, using the quotient rule and the chain rule

$T'(t) = ( -2t cos(t^2) , -2t sin(t^2))$

And also

$|T'(t)| = \sqrt{ ( -2t cos(t^2))^2 + (-2t sin(t^2))^2} = \sqrt{ 4t^2 ( ( cos(t^2))^2 + ( sin(t^2))^2)} = \sqrt{4t^2} \\ = 2t$

Therefore

$N(t) = T'(t) / |T'(t) | = (-cos(t^2) , -sin(t^2))$

Notice that

1. $|N(t)| = |T(t) | = \sqrt{ cos(t^2)^2 + sin(t^2)^2 } = \sqrt{1} = 1$

2. $N(t)*T(T) = cos(t^2) sin(t^2 ) - cos(t^2) sin(t^2 ) = 0$

Simlarly

$r'(t) = (2t,-6,0) \\$

and

$|r'(t)| = \sqrt{(2t)^2 + 6^2} = \sqrt{4t^2 + 36}$

Therefore

$T(t) =r'(t)/|r'(t)| = (2t/ \sqrt{4t^2 + 36} , -6/\sqrt{4t^2 + 36},0)$

Then

$T'(t) = (9/(9 + t^2)^{3/2} , (3 t)/(9 + t^2)^{3/2},0)$

and also

$|T'(t)| = \sqrt{ ( (9/(9 + t^2)^{3/2} )^2 + ( (3 t)/(9 + t^2)^{3/2})^2 + 0^2 }\\= 3/(t^2 + 9 )$

And since

$N(t) =T(t)/|T'(t)| = (3/(9 + t^2)^{1/2} , t/(9 + t^2)^{1/2},0)$

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Yall just use me 4 points }:

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Read 2 more answers

If tan tetha =8/15,find the value of sin tetha+cos tetha all divided by cos tetha (1-cos tetha)​

If tan tetha =8/15,find the value of sin tetha+cos tetha all divided by cos tetha (1-cos tetha)