Step-by-step explanation:
![\mathsf{Given :\;\dfrac{{sec}^2\theta - co{sec}^2\theta}{{sec}^2\theta + co{sec}^2\theta}}](https://tex.z-dn.net/?f=%5Cmathsf%7BGiven%20%3A%5C%3B%5Cdfrac%7B%7Bsec%7D%5E2%5Ctheta%20-%20co%7Bsec%7D%5E2%5Ctheta%7D%7B%7Bsec%7D%5E2%5Ctheta%20%2B%20co%7Bsec%7D%5E2%5Ctheta%7D%7D)
![\bigstar\;\;\textsf{We know that : \large\boxed{\mathsf{{sec}\theta = \dfrac{1}{cos\theta}}}}](https://tex.z-dn.net/?f=%5Cbigstar%5C%3B%5C%3B%5Ctextsf%7BWe%20know%20that%20%3A%20%5Clarge%5Cboxed%7B%5Cmathsf%7B%7Bsec%7D%5Ctheta%20%3D%20%5Cdfrac%7B1%7D%7Bcos%5Ctheta%7D%7D%7D%7D)
![\bigstar\;\;\textsf{We know that : \large\boxed{\mathsf{co{sec}\theta = \dfrac{1}{sin\theta}}}}](https://tex.z-dn.net/?f=%5Cbigstar%5C%3B%5C%3B%5Ctextsf%7BWe%20know%20that%20%3A%20%5Clarge%5Cboxed%7B%5Cmathsf%7Bco%7Bsec%7D%5Ctheta%20%3D%20%5Cdfrac%7B1%7D%7Bsin%5Ctheta%7D%7D%7D%7D)
![\mathsf{\implies \dfrac{\dfrac{1}{cos^2\theta} - \dfrac{1}{sin^2\theta}}{\dfrac{1}{cos^2\theta} + \dfrac{1}{sin^2\theta}}}](https://tex.z-dn.net/?f=%5Cmathsf%7B%5Cimplies%20%5Cdfrac%7B%5Cdfrac%7B1%7D%7Bcos%5E2%5Ctheta%7D%20-%20%5Cdfrac%7B1%7D%7Bsin%5E2%5Ctheta%7D%7D%7B%5Cdfrac%7B1%7D%7Bcos%5E2%5Ctheta%7D%20%2B%20%5Cdfrac%7B1%7D%7Bsin%5E2%5Ctheta%7D%7D%7D)
![\mathsf{\implies \dfrac{\dfrac{sin^2\theta - cos^2\theta}{sin^2\theta.cos^2\theta}}{\dfrac{sin^2\theta + cos^2\theta}{sin^2\theta.cos^2\theta}}}](https://tex.z-dn.net/?f=%5Cmathsf%7B%5Cimplies%20%5Cdfrac%7B%5Cdfrac%7Bsin%5E2%5Ctheta%20-%20cos%5E2%5Ctheta%7D%7Bsin%5E2%5Ctheta.cos%5E2%5Ctheta%7D%7D%7B%5Cdfrac%7Bsin%5E2%5Ctheta%20%2B%20cos%5E2%5Ctheta%7D%7Bsin%5E2%5Ctheta.cos%5E2%5Ctheta%7D%7D%7D)
![\mathsf{\implies \dfrac{sin^2\theta - cos^2\theta}{sin^2\theta + cos^2\theta}}](https://tex.z-dn.net/?f=%5Cmathsf%7B%5Cimplies%20%5Cdfrac%7Bsin%5E2%5Ctheta%20-%20cos%5E2%5Ctheta%7D%7Bsin%5E2%5Ctheta%20%2B%20cos%5E2%5Ctheta%7D%7D)
Taking sin²θ common in both numerator & denominator, We get :
![\mathsf{\implies \dfrac{sin^2\theta\left(1 - \dfrac{cos^2\theta}{sin^2\theta}\right)}{sin^2\theta\left(1 + \dfrac{cos^2\theta}{sin^2\theta}\right)}}](https://tex.z-dn.net/?f=%5Cmathsf%7B%5Cimplies%20%5Cdfrac%7Bsin%5E2%5Ctheta%5Cleft%281%20-%20%5Cdfrac%7Bcos%5E2%5Ctheta%7D%7Bsin%5E2%5Ctheta%7D%5Cright%29%7D%7Bsin%5E2%5Ctheta%5Cleft%281%20%2B%20%5Cdfrac%7Bcos%5E2%5Ctheta%7D%7Bsin%5E2%5Ctheta%7D%5Cright%29%7D%7D)
![\bigstar\;\;\textsf{We know that : \large\boxed{\mathsf{cot\theta = \dfrac{cos\theta}{sin\theta}}}}](https://tex.z-dn.net/?f=%5Cbigstar%5C%3B%5C%3B%5Ctextsf%7BWe%20know%20that%20%3A%20%5Clarge%5Cboxed%7B%5Cmathsf%7Bcot%5Ctheta%20%3D%20%5Cdfrac%7Bcos%5Ctheta%7D%7Bsin%5Ctheta%7D%7D%7D%7D)
![\mathsf{\implies \dfrac{1 -cot^2\theta}{1 + cot^2\theta}}](https://tex.z-dn.net/?f=%5Cmathsf%7B%5Cimplies%20%5Cdfrac%7B1%20-cot%5E2%5Ctheta%7D%7B1%20%2B%20cot%5E2%5Ctheta%7D%7D)
![\mathsf{Given :\;cot\theta = \dfrac{1}{\sqrt{5}}}](https://tex.z-dn.net/?f=%5Cmathsf%7BGiven%20%3A%5C%3Bcot%5Ctheta%20%3D%20%5Cdfrac%7B1%7D%7B%5Csqrt%7B5%7D%7D%7D)
![\mathsf{\implies \dfrac{1 - \left(\dfrac{1}{\sqrt{5}}\right)^2}{1 + \left(\dfrac{1}{\sqrt{5}}\right)^2}}](https://tex.z-dn.net/?f=%5Cmathsf%7B%5Cimplies%20%5Cdfrac%7B1%20-%20%5Cleft%28%5Cdfrac%7B1%7D%7B%5Csqrt%7B5%7D%7D%5Cright%29%5E2%7D%7B1%20%2B%20%5Cleft%28%5Cdfrac%7B1%7D%7B%5Csqrt%7B5%7D%7D%5Cright%29%5E2%7D%7D)
![\mathsf{\implies \dfrac{1 - \dfrac{1}{5}}{1 + \dfrac{1}{5}}}](https://tex.z-dn.net/?f=%5Cmathsf%7B%5Cimplies%20%5Cdfrac%7B1%20-%20%5Cdfrac%7B1%7D%7B5%7D%7D%7B1%20%2B%20%5Cdfrac%7B1%7D%7B5%7D%7D%7D)
![\mathsf{\implies \dfrac{\dfrac{5 - 1}{5}}{\dfrac{5 + 1}{5}}}](https://tex.z-dn.net/?f=%5Cmathsf%7B%5Cimplies%20%5Cdfrac%7B%5Cdfrac%7B5%20-%201%7D%7B5%7D%7D%7B%5Cdfrac%7B5%20%2B%201%7D%7B5%7D%7D%7D)
![\mathsf{\implies \dfrac{5 - 1}{5 + 1}}](https://tex.z-dn.net/?f=%5Cmathsf%7B%5Cimplies%20%5Cdfrac%7B5%20-%201%7D%7B5%20%2B%201%7D%7D)
![\mathsf{\implies \dfrac{4}{6}}](https://tex.z-dn.net/?f=%5Cmathsf%7B%5Cimplies%20%5Cdfrac%7B4%7D%7B6%7D%7D)
![\mathsf{\implies \dfrac{2}{3}}](https://tex.z-dn.net/?f=%5Cmathsf%7B%5Cimplies%20%5Cdfrac%7B2%7D%7B3%7D%7D)
<u>Hence</u><u>,</u><u> option</u><u> </u><u>(</u><u>a)</u><u> </u><u>2</u><u>/</u><u>3</u><u> </u><u>is </u><u>your</u><u> </u><u>correct</u><u> </u><u>answer</u><u>.</u>
The answer is shout is to whisper
explanation:
answer
Answer:
g(x)=x-3
Problem:
If f(x+3)=x+6 find inverse of function f(x).
Step-by-step explanation:
Let u=x+3, then x=u-3.
Make this substitution into our given:
f(u)=(u-3)+6
Simplify:
f(u)=u+(-3+6)
f(u)=u+3
Now let's find the inverse of f(u)...
Or if you prefer rename the variable...
f(x)=x+3
Now, we are going to solve y=x+3 for x.
Subtracting 3 on both sides gives: y-3=x.
Interchange x and y: x-3=y.
So the inverse of f(x)=x+3 is g(x)=x-3.
9+12>18
12+18>9
9+18>12
So, yes. A triangle can be formed by the given angles.
Answer:A=P{1+r/n}^nt. A is the end value, P the start. r is the rate, divided by n, the number of compoundings per year. t is the number of years.
semiannual is 2 compoundings
A=7000{1+(0.055/2)}^6, the 6 from 2 compoundings a year for 3 years. Do the parentheses first, so that you have 1.0.0257^6. When you get that, multiply by $7000. Round at the end, not intermediate steps.
A=$8237.38
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A=7000{1+(.055/4)}^12=$8246.48. Notice that what is in the parentheses gets smaller, but it gets raised to a higher power.
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A=7000{1+(0.055/12)}^36=$8252.64