Please tell
Asked by | 20th Jan, 2009, 09:08: PM
Expert Answer:
3sinA - 4cosA = 5
(3/5)sinA - (4/5)cosA = 1
let cosB = 3/5 then sinB = 4/5
cosBsinA - sinBcosA = sin 90
sin(A-B) = sin 90
A-B = 90
A = 90+B
now 3cosA+4sinA
=3cos(90+B) + 4sin(90+B)
=3(-sinB) + 4cosB
=3(-4/5) + 4(3/5)
=0
Answered by | 22nd Jan, 2009, 10:21: AM
Related Videos
Kindly Sign up for a personalised experience
- Ask Study Doubts
- Sample Papers
- Past Year Papers
- Textbook Solutions
Sign Up
Verify mobile number
Enter the OTP sent to your number
Change