Cos 8A. COS 5A-COS 12A COS9A/
Sin 8A COSSA + Cos 12A . Sin 9A
Answers
Answer:
(iii) (p^2 - q^2r)^2+ 2p^2q^2r
Step-by-step explanation:
(iii) (p^2 - q^2r)^2+ 2p^2q^2r
(iii) (p^2 - q^2r)^2+ 2p^2q^2r
(iii) (p^2 - q^2r)^2+ 2p^2q^2r
(iii) (p^2 - q^2r)^2+ 2p^2q^2r
Answer:
2cos8A cos5A-2cos12A cos9A/2sin 8A cos 5A+cos 12Asin 9A
= Cos 13A+Cos 3A-cos21A-cos3A/ Sin13A+sin3A+Sin 21A-sin 3A
= Cos13A-cos21A/Sin13A+sin 21A
= 2sin12Asin4A/2sin12Acos4A
= Tan 4A
Step-by-step explanation:
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