Cos alpha+cos beta=0=sin alpha +sin beta then find cos (alpha-beta)
Cos alpha+cos beta=0=sin alpha +sin beta then find cos (alpha-beta)
Gayathri , 6 Years ago
Grade 11
Trigonometry> Cos alpha+cos beta=0=sin alpha +sin beta ...
1 AnswersSiddhartha reddy
Last Activity: 6 Years ago
Given,
Cos alpha + cos beta = 0, cos alpha = -cos beta
Sin alpha + sin beta = 0, sin alpha = -sin beta
Cos(alpha-beta) = cos alpha × cos beta + sin alpha × sin beta
= - cos beta × cos beta + (-sin beta ) × sin beta
= -cos^2 beta - sin^2 beta
=-(cos^2 beta + sin^2 beta)
=-(1)
=-1
LIVE ONLINE CLASSES
Prepraring for the competition made easy just by live online class.
Full Live Access
Study Material
Live Doubts Solving
Daily Class Assignments
Other Related Questions on trigonometry
In ABC, if cot A+cot B+cot C = cosecA cosecB cosecC, then prove that ABC is right angle triangle.
In ABC, if cot A+cot B+cot C = cosecA cosecB cosecC, then prove that ABC is right angle triangle.
trigonometry
1 Answer Available
Last Activity: 2 Years ago
- Find the general solution of equation tan3∂=cot2∂
For all angles between -720 and +720
- Find the general solution of equation tan3∂=cot2∂
For all angles between -720 and +720
trigonometry
0 Answer Available
Last Activity: 3 Years ago
To prove : Cosec(45 degree - A)Cosec(45 degree + A) = 2SecA
To prove : Cosec(45 degree - A)Cosec(45 degree + A) = 2SecA
trigonometry
0 Answer Available
Last Activity: 3 Years ago
cos^2 13ºcos^2 47º+cos^2 73º =? …..................
cos^2 13ºcos^2 47º+cos^2 73º =? …..................
trigonometry
0 Answer Available
Last Activity: 3 Years ago
Derive a formula for the area of the triangle, given b, A, B, and C. That is, derive the formula: (b^2sinAsinC)/(2sinB)
Derive a formula for the area of the triangle, given b, A, B, and C. That is, derive the formula: (b^2sinAsinC)/(2sinB)
trigonometry
0 Answer Available
Last Activity: 3 Years ago