We start with the more complicated side which is the left side, and show that, on using some trigonometric identities, we will get the term on the right side .
![\frac{sin \theta + tan \theta}{1+cos \theta}](https://tex.z-dn.net/?f=%5Cfrac%7Bsin%20%5Ctheta%20%2B%20tan%20%5Ctheta%7D%7B1%2Bcos%20%5Ctheta%7D)
Using Quotient identity for tangent function, we will get
![\frac{sin \theta+ \frac{sin \theta}{cos \theta}}{1+cos \theta}](https://tex.z-dn.net/?f=%5Cfrac%7Bsin%20%5Ctheta%2B%20%5Cfrac%7Bsin%20%5Ctheta%7D%7Bcos%20%5Ctheta%7D%7D%7B1%2Bcos%20%5Ctheta%7D)
![\frac{sin \theta cos \theta + sin \theta}{cos \theta(1+cos \theta)}](https://tex.z-dn.net/?f=%5Cfrac%7Bsin%20%5Ctheta%20cos%20%5Ctheta%20%2B%20sin%20%5Ctheta%7D%7Bcos%20%5Ctheta%281%2Bcos%20%5Ctheta%29%7D)
Taking out sine function from the numerator
![=\frac{sin \theta(1+cos \theta)}{cos \theta(1+cos \theta)}](https://tex.z-dn.net/?f=%3D%5Cfrac%7Bsin%20%5Ctheta%281%2Bcos%20%5Ctheta%29%7D%7Bcos%20%5Ctheta%281%2Bcos%20%5Ctheta%29%7D)
Cancelling the common term of numerator and denominator
![=\frac{sin \theta}{cos \theta} = tan \theta](https://tex.z-dn.net/?f=%3D%5Cfrac%7Bsin%20%5Ctheta%7D%7Bcos%20%5Ctheta%7D%20%3D%20tan%20%5Ctheta)
Answer:
80 cubes
Step-by-step explanation:
volume of one small cube = 1 /2 × 1 / 2 × 1 / 2
= 1 / 8cm^3
let x be the number of cubes needed
1 / 8 × x = 10
x = 10 × 8
x = 80 cubes
Answer:
Step-by-step explanation:
It asks for the common set of A and B.
<u>There only one element common to both the given sets:</u>
- A={2,3,5,7} and B={2,4,6,8} ⇒ A∩B = {2}
Answer:
1, 2, 8, 128, 32768
Step-by-step explanation:
![recursive \: formula: \\ a _{n} = 2. {( a_{n - 1} )}^{2} \\ \\ a _{1} = 1...(given) \\ \\ a _{2} = 2. {( a_{2 - 1} )}^{2} = 2 {(a _{1} )}^{2} = 2 {(1)}^{2} \\ \\ a _{2} = 2\\ \\a _{3} = 2. {( a_{3 - 1} )}^{2} = 2 {(a _{2} )}^{2} = 2 {(2)}^{2} \\ \\ a _{3} = 8\\ \\a _{4} = 2. {( a_{4 - 1} )}^{2} = 2 {(a _{3} )}^{2} = 2 {(8)}^{2} \\ \\ a _{4} = 128\\ \\a _{5} = 2. {( a_{5 - 1} )}^{2} = 2 {(a _{4} )}^{2} = 2 {(128)}^{2} \\ \\ a _{5} = 32,768\\ \\](https://tex.z-dn.net/?f=recursive%20%5C%3A%20formula%3A%20%5C%5C%20%20a%20_%7Bn%7D%20%3D%202.%20%7B%28%20a_%7Bn%20-%201%7D%20%29%7D%5E%7B2%7D%20%20%5C%5C%20%20%5C%5C%20a%20_%7B1%7D%20%3D%201...%28given%29%20%5C%5C%20%20%5C%5C%20a%20_%7B2%7D%20%3D%202.%20%7B%28%20a_%7B2%20-%201%7D%20%29%7D%5E%7B2%7D%20%20%20%3D%202%20%7B%28a%20_%7B1%7D%20%29%7D%5E%7B2%7D%20%20%3D%202%20%7B%281%29%7D%5E%7B2%7D%20%20%5C%5C%20%20%5C%5C%20%20a%20_%7B2%7D%20%3D%202%5C%5C%20%20%5C%5Ca%20_%7B3%7D%20%3D%202.%20%7B%28%20a_%7B3%20-%201%7D%20%29%7D%5E%7B2%7D%20%20%20%3D%202%20%7B%28a%20_%7B2%7D%20%29%7D%5E%7B2%7D%20%20%3D%202%20%7B%282%29%7D%5E%7B2%7D%20%20%5C%5C%20%20%5C%5C%20%20a%20_%7B3%7D%20%3D%208%5C%5C%20%20%5C%5Ca%20_%7B4%7D%20%3D%202.%20%7B%28%20a_%7B4%20-%201%7D%20%29%7D%5E%7B2%7D%20%20%20%3D%202%20%7B%28a%20_%7B3%7D%20%29%7D%5E%7B2%7D%20%20%3D%202%20%7B%288%29%7D%5E%7B2%7D%20%20%5C%5C%20%20%5C%5C%20%20a%20_%7B4%7D%20%3D%20128%5C%5C%20%20%5C%5Ca%20_%7B5%7D%20%3D%202.%20%7B%28%20a_%7B5%20-%201%7D%20%29%7D%5E%7B2%7D%20%20%20%3D%202%20%7B%28a%20_%7B4%7D%20%29%7D%5E%7B2%7D%20%20%3D%202%20%7B%28128%29%7D%5E%7B2%7D%20%20%5C%5C%20%20%5C%5C%20%20a%20_%7B5%7D%20%3D%2032%2C768%5C%5C%20%20%5C%5C)
Thus the first five terms are: 1, 2, 8, 128, 32768.
Answer:
Option (d) is correct.
![\sqrt{10000x^{64}}=100x^{32}](https://tex.z-dn.net/?f=%5Csqrt%7B10000x%5E%7B64%7D%7D%3D100x%5E%7B32%7D)
Step-by-step explanation:
Given : Expression ![\sqrt{10000x^{64}}](https://tex.z-dn.net/?f=%5Csqrt%7B10000x%5E%7B64%7D%7D)
We have to write a simplified form of the given expression ![\sqrt{10000x^{64}}](https://tex.z-dn.net/?f=%5Csqrt%7B10000x%5E%7B64%7D%7D)
Consider the given expression ![\sqrt{10000x^{64}}](https://tex.z-dn.net/?f=%5Csqrt%7B10000x%5E%7B64%7D%7D)
![\mathrm{Apply\:radical\:rule\:}\sqrt[n]{ab}=\sqrt[n]{a}\sqrt[n]{b},\:\quad \mathrm{\:assuming\:}a\ge 0,\:b\ge 0](https://tex.z-dn.net/?f=%5Cmathrm%7BApply%5C%3Aradical%5C%3Arule%5C%3A%7D%5Csqrt%5Bn%5D%7Bab%7D%3D%5Csqrt%5Bn%5D%7Ba%7D%5Csqrt%5Bn%5D%7Bb%7D%2C%5C%3A%5Cquad%20%5Cmathrm%7B%5C%3Aassuming%5C%3A%7Da%5Cge%200%2C%5C%3Ab%5Cge%200)
![=\sqrt{10000}\sqrt{x^{64}}](https://tex.z-dn.net/?f=%3D%5Csqrt%7B10000%7D%5Csqrt%7Bx%5E%7B64%7D%7D)
Factor 10000 as ![10000=100^2](https://tex.z-dn.net/?f=10000%3D100%5E2)
![\mathrm{Apply\:radical\:rule}:\quad \sqrt[n]{a^n}=a](https://tex.z-dn.net/?f=%5Cmathrm%7BApply%5C%3Aradical%5C%3Arule%7D%3A%5Cquad%20%5Csqrt%5Bn%5D%7Ba%5En%7D%3Da)
![\sqrt{100^2}=100](https://tex.z-dn.net/?f=%5Csqrt%7B100%5E2%7D%3D100)
also, ![\mathrm{Apply\:radical\:rule\:}\sqrt[n]{a^m}=a^{\frac{m}{n}},\:\quad \mathrm{\:assuming\:}a\ge 0](https://tex.z-dn.net/?f=%5Cmathrm%7BApply%5C%3Aradical%5C%3Arule%5C%3A%7D%5Csqrt%5Bn%5D%7Ba%5Em%7D%3Da%5E%7B%5Cfrac%7Bm%7D%7Bn%7D%7D%2C%5C%3A%5Cquad%20%5Cmathrm%7B%5C%3Aassuming%5C%3A%7Da%5Cge%200)
We have,
![\sqrt{x^{64}}=x^{\frac{64}{2}}=x^{32}](https://tex.z-dn.net/?f=%5Csqrt%7Bx%5E%7B64%7D%7D%3Dx%5E%7B%5Cfrac%7B64%7D%7B2%7D%7D%3Dx%5E%7B32%7D)
Thus, ![\sqrt{10000x^{64}}=100x^{32}](https://tex.z-dn.net/?f=%5Csqrt%7B10000x%5E%7B64%7D%7D%3D100x%5E%7B32%7D)