PLEASE SOLVE THIS AND HELP ME OUT
Answers
Answer:
Given, (cos 8A * cos 5A - cos 12A * cos 9A)/(sin 8A * cos 5A + cos 12A * sin 9A)
= {2(cos 8A * cos 5A - cos 12A * cos9A)/2}/{2(sin 8A * cos 5A + cos 12A * sin 9A)/2} {Multiply and divide by 2}
= [{cos 13A + cos 3A}/2 - {cos 21A + cos 3A}/2]/[{sin 13A + sin 3A}/2 + {sin 21A - sin 3A}/2] {Apply product into sum or difference formula of
trigonometric ratios}
= (cos 13A - cos 21A)/(sin 13A + sin 21A)
= (2*sin 17A * sin 4A)/(2*sin 17A * cos4A) {Apply sum and difference into product formula of trigonometric ratios}
= sin 4A/cos 4A
= tan 4A
So, (cos 8A * cos 5A - cos 12A * cos 9A)/(sin 8A * cos 5A + cos 12A * sin 9A) = tan 4A
hope it's help u......
Step-by-step explanation:
Use the Transformation formulae
2cosAcosB=cos(A+B)+cos(A-B) and 2sinAcosB=sin(A+B) + sin(A-B)