Question

PLEASE SOLVE THIS AND HELP ME OUT

Answer 1

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......

Answer 2

Step-by-step explanation:

Use the Transformation formulae

2cosAcosB=cos(A+B)+cos(A-B) and 2sinAcosB=sin(A+B) + sin(A-B)

$=\frac{cos8Acos5A-cos12Acos9A}{sin8Acos5A+cos12Asin9A}\\=\frac{cos13A+cos3A-cos21A-cos3A}{sin13A+sin3A+sin21A-sin3A}\\=\frac{cos13A-cos21A}{sin13A+sin21A}$

$=\frac{-2sin17Asin(-4A)}{2sin17Acos4A}\\\\=tan4A$