Answer:
sine and cosec are inverse of each other.
cosine and sec are inverse of each other.
tan and cot are inverse of each other.
Step-by-step explanation:
Given point on terminal side of an angle (
).
Kindly refer to the attached image for the diagram of the given point.
Let it be point A(
)
Let O be the origin i.e. (0,0)
Point B will be (
)
Now, let us consider the right angled triangle
:
Sides:
![Base, OB = \frac{1}{3}\\Perpendicular, AB = \frac{1}{4}](https://tex.z-dn.net/?f=Base%2C%20OB%20%3D%20%5Cfrac%7B1%7D%7B3%7D%5C%5CPerpendicular%2C%20AB%20%3D%20%5Cfrac%7B1%7D%7B4%7D)
Using Pythagorean theorem:
![\text{Hypotenuse}^{2} = \text{Base}^{2} + \text{Perpendicular}^{2}\\\Rightarrow OA^{2} = OB^{2} + AB^{2}\\\Rightarrow OA^{2} = \frac{1}{3}^{2} + \frac{1}{4}^{2}\\\Rightarrow OA = \sqrt{\frac{1}{3}^{2} + \frac{1}{4}^{2}}\\\Rightarrow OA = \sqrt{\frac{4^2+3^2}{3^{2}.4^2 }}\\\Rightarrow OA = \frac{5}{12}](https://tex.z-dn.net/?f=%5Ctext%7BHypotenuse%7D%5E%7B2%7D%20%3D%20%5Ctext%7BBase%7D%5E%7B2%7D%20%2B%20%5Ctext%7BPerpendicular%7D%5E%7B2%7D%5C%5C%5CRightarrow%20OA%5E%7B2%7D%20%3D%20OB%5E%7B2%7D%20%2B%20AB%5E%7B2%7D%5C%5C%5CRightarrow%20OA%5E%7B2%7D%20%3D%20%5Cfrac%7B1%7D%7B3%7D%5E%7B2%7D%20%2B%20%5Cfrac%7B1%7D%7B4%7D%5E%7B2%7D%5C%5C%5CRightarrow%20OA%20%3D%20%5Csqrt%7B%5Cfrac%7B1%7D%7B3%7D%5E%7B2%7D%20%2B%20%5Cfrac%7B1%7D%7B4%7D%5E%7B2%7D%7D%5C%5C%5CRightarrow%20OA%20%3D%20%5Csqrt%7B%5Cfrac%7B4%5E2%2B3%5E2%7D%7B3%5E%7B2%7D.4%5E2%20%7D%7D%5C%5C%5CRightarrow%20OA%20%3D%20%5Cfrac%7B5%7D%7B12%7D)
![sin \angle AOB = \dfrac{Perpendicular}{Hypotenuse}](https://tex.z-dn.net/?f=sin%20%5Cangle%20AOB%20%3D%20%5Cdfrac%7BPerpendicular%7D%7BHypotenuse%7D)
![\Rightarrow sin \angle AOB = \dfrac{\frac{1}{4}}{\frac{5}{12}}\\\Rightarrow sin \angle AOB = \dfrac{3}{5}](https://tex.z-dn.net/?f=%5CRightarrow%20sin%20%5Cangle%20AOB%20%3D%20%5Cdfrac%7B%5Cfrac%7B1%7D%7B4%7D%7D%7B%5Cfrac%7B5%7D%7B12%7D%7D%5C%5C%5CRightarrow%20sin%20%5Cangle%20AOB%20%3D%20%5Cdfrac%7B3%7D%7B5%7D)
![cos\angle AOB = \dfrac{Base}{Hypotenuse}](https://tex.z-dn.net/?f=cos%5Cangle%20AOB%20%3D%20%5Cdfrac%7BBase%7D%7BHypotenuse%7D)
![\Rightarrow cos \angle AOB = \dfrac{\frac{1}{3}}{\frac{5}{12}}\\\Rightarrow cos\angle AOB = \dfrac{4}{5}](https://tex.z-dn.net/?f=%5CRightarrow%20cos%20%5Cangle%20AOB%20%3D%20%5Cdfrac%7B%5Cfrac%7B1%7D%7B3%7D%7D%7B%5Cfrac%7B5%7D%7B12%7D%7D%5C%5C%5CRightarrow%20cos%5Cangle%20AOB%20%3D%20%5Cdfrac%7B4%7D%7B5%7D)
![tan\angle AOB = \dfrac{Perpendicular}{Base}](https://tex.z-dn.net/?f=tan%5Cangle%20AOB%20%3D%20%5Cdfrac%7BPerpendicular%7D%7BBase%7D)
![\Rightarrow tan\angle AOB = \dfrac{3}{4}](https://tex.z-dn.net/?f=%5CRightarrow%20tan%5Cangle%20AOB%20%3D%20%5Cdfrac%7B3%7D%7B4%7D)
![cosec \angle AOB = \dfrac{Hypotenuse}{Perpendicular}](https://tex.z-dn.net/?f=cosec%20%5Cangle%20AOB%20%3D%20%5Cdfrac%7BHypotenuse%7D%7BPerpendicular%7D)
![\Rightarrow cosec\angle AOB = \dfrac{5}{3}](https://tex.z-dn.net/?f=%5CRightarrow%20cosec%5Cangle%20AOB%20%3D%20%5Cdfrac%7B5%7D%7B3%7D)
![sec\angle AOB = \dfrac{Hypotenuse}{Base}](https://tex.z-dn.net/?f=sec%5Cangle%20AOB%20%3D%20%5Cdfrac%7BHypotenuse%7D%7BBase%7D)
![\Rightarrow sec\angle AOB = \dfrac{5}{4}](https://tex.z-dn.net/?f=%5CRightarrow%20sec%5Cangle%20AOB%20%3D%20%5Cdfrac%7B5%7D%7B4%7D)
![cot\angle AOB = \dfrac{Base}{Perpendicular}](https://tex.z-dn.net/?f=cot%5Cangle%20AOB%20%3D%20%5Cdfrac%7BBase%7D%7BPerpendicular%7D)
![\Rightarrow cot\angle AOB = \dfrac{4}{3}](https://tex.z-dn.net/?f=%5CRightarrow%20cot%5Cangle%20AOB%20%3D%20%5Cdfrac%7B4%7D%7B3%7D)
<span>instead of showing that the conclusion to be proved is true, you show that all of the alternatives are false. To do this, you must assume the negation of the statement to be proved. Then, deductive reasoning will lead to a contradiction: two statements that cannot both be true.</span>
Answer:
Step-by-step explanation:
Answer:
71/44
Step-by-step explanation:
Here the LCD is the product of the two denominators given: 11*8 = 88.
We rewrite 4/11 as 32/88 and 10/8 as 110/88. Now the denominators are the same.
Combine 32/88 and 110/88. We get the sum 142/88, which reduces to 71/44.
A=17 b = 12
a x b
17 x 12
=204