ab3n — b3mn—a2b2+ab3m=a2bm- - a2b2 —— a2mn+a2bn; - .. (a2—b2) mn=a2bm—ab2m+a2bn—ab2n =ab (am-bman-bn) .. (a+b) mn=ab (m+n), (4) If m=cosec A-sin A; and n=sec A cos A; (7) tan A+cot A=2, find A. A=45°. (8) Write down the sum of all the exterior and also of all the interior angles of a polygon of n sides; and thence deduce the value of each of the vertical angles of the triangles made by producing both ways all the sides of a regular polygon, and verify the result in the cases where n=3 and n=4. (12) sin1 A-2 sin2 4-1-2 sin 4-cos2 4, find A. A=270°. (13) tan1 4-tan2 4+1=4 tan A+sec2 4, find tan A. Find tan x and tan y, without quadratics, (24) (tan A-1)-8 cot (1-2 cot 4)=8: find tan A. (25) Prove that 1 1 tan A+tan A÷&c., ad inf. ='{{√(sec2 4+3)—tan 4}. CHAPTER II, 7. To find the sine and cosine of the sum and difference of two angles. B Р F G C Let the angle FAC=A, and the angle FAP=B, then PAC=A+B, draw PF perpendicular to AF, and from P and F let fall the perpendiculars PD and FE on AC, and draw FG parallel to AC. Now, because GF is parallel to AC, the alternate angles, AFG and FAC are equal, A but FAC=A, therefore AFG=A; and since AFP is a right angle, the angle PFG is the complement of AFG; but FGP is a right angle, therefore FPG is the complement of PFG, that is, the angles FPG and AFG are the complements of the same angle (PFG), they are therefore equal; but AFG has been proved equal to A, hence FPG=A. DP DG+GP EF+PG DE sin (A+B)= = .. sin (A+B)=sin A cos B+cos A sin B. AD AE-DE AE GF cos (A+B)= AP AP AP AP' (because GFDE) F P To find the sine and cosine of A-B. Let the angle DAF-A, and the angle PAF-B, then ▲ PAD=A—B: from any point P draw the perpendiculars PF and PD on AF and AD, also draw PG parallel to AD, then PFG the complement of AFE, and therefore A, because AFG is the complement of A; and because GD is a rectangle, ED PG, and PD=GE. E D 8. Now, since sin (A+B)=sin A cos B+cos A sin B, if we make B=A, then sin(4+4)=sin 2 4= sin A cos 4+cos A sin 4-2 sin cosД; also, since cos (A+B)=cos A cos B—sin A sin B ; make B=A; then cos (4+4)=cos 2 A= cos A cos A—sin A sin A cos2 4-sin2 A; but since cos A-1-sin A, we bave also cos 2 A-1-sin2 A-sin2 4=1 -2 sin2 A; also sin2 4=1-cos2 A; .. cos 2 Acos2 A-(1-cos2 A)=2 cos2 A-1. Since A may be any angle, we have the sine of any angle equal to twice the sine of half that angle multiplied by the cosine of half that angle; and the cosine of any angle is equal to the square of the cosine of half that angle minus the sine square of half that angle. For the sines we have |