Question

prove that Cos8A.cos5A-cos12A.cos9A / sin8A.cos5A+cos12A.sin9A=tan4a

Answer 1

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

Answer 2

Question :--

(cos8A.cos5A-cos12A.cos9A)/(sin8A.cos5A+cos12A.sin9A)

solution :----

we know that,

cosA*cosB = 1/2[(cos(A+B)+cos(A-B)

sinA*cosB = 1/2[sin(A+B)+sin(A-B)

cosA*sinB = 1/2[sin(A+B) - sin(A-B)

using all values , we get,

= [1/2 {cos13A+cos3A}-1/2 {cos21A+cos3A}]/[1/2 {sin13A+sin3A}+1/2 {sin21A-sin3A}]

= ( cos13A-cos21A)/(sin13A+sin21A)

= 2sin17A.sin4A/2sin17A.cos4A

= tan4A $\huge\boxed{\fcolorbox{fuchsia}{grey}{Proved}}$