Answer:
multiply the first equation with 3, then add the two equations, you'll get the value of x, by putting the value of x in any of the two equations, find y, answer should be this,
x = 33/13,
y = 18/13
1+tan^2=sec^2
so, given expression=
![\frac{sec^{2}*cot}{cosec^{2} } = \frac{ \frac{1*cot}{ cos^{2} } }{ \frac{1}{ sin^{2} } } =](https://tex.z-dn.net/?f=%20%5Cfrac%7Bsec%5E%7B2%7D%2Acot%7D%7Bcosec%5E%7B2%7D%20%7D%20%3D%20%5Cfrac%7B%20%5Cfrac%7B1%2Acot%7D%7B%20cos%5E%7B2%7D%20%7D%20%7D%7B%20%5Cfrac%7B1%7D%7B%20sin%5E%7B2%7D%20%7D%20%7D%20%3D)
![tan^{2}*cot = tan ](https://tex.z-dn.net/?f=%20tan%5E%7B2%7D%2Acot%20%3D%20tan%20%0A)
(using sin/cos=tan and tan*cot=1)
hence proved
Answer:
1856.37 m
Step-by-step explanation:
When the tapes were taken as 30.0 m long, the measured distance was 1858.23 m. This is equivalent to
tapes.
With the correct tape length now 29.97 m, the measured distance is
to 2 decimal places.
Answer: im not sure. that looks like 10th grade work
Step-by-step explanation: