. + tan 3 7) 54. tan A + tan B = sec A . sec B. sin (A + B). А B 2 sin A 2 cos A + cos B' 59. If tan A s and tan B find tan (A + B) and tan (A - B). 60. If tan A = 1 and tan B }, show that tan (A + B) = 1. 61. If (1+tan A) (1 + tan B)= 2, show that tan (A + B) = 1. 62. Show that if 45° is divided into two parts, 0 and 0, then (1 + tan 0) (1 + tan o)= 2. 63. tan A tan 2 A tan 3A = tan 3A – tan 2A – tan A. 64. If tan 6=2m+1 and tan = 2m-1, then cot (0-4)= 2m”. 65. sin (A + B). sin (4 – B) = sino4 – sin’B= cos B - cos’A. 66. cos (A + B). cos (A – B) = cos A – sin’B= cosB – sin’A. = . 73. cos 20 + 3 sin A = 2. 74. tan 6 + tan 20 + 1/3 tan 0. tan 20 = /3. 75. 13 (tan 0 + tan 20) + tan 8. tan 20 = 1. 76. cos 20 – sin 0 = 1 136. We have at present defined the ratios of acute angles only. We now proceed to define those of obtuse angles. Then I'OF is an acute angle. In or take any point B: and draw BH at right angles to Ol' meeting OF in H. = OH , = Then BOH is the right-angled triangle by which we determine the ratios of the acute angle I'OF, OB being base, OH being hypothenuse, and BH being perpendicular, Now it is found convenient to use this same right-angled triangle BOH to define the ratios of the obtuse angle 10 F; only that, as OB has to be drawn in the opposite direction to the line 01, which bounds the obtuse angle, we take – OB (instead of OB) as its base. Thus BH – ОВ cos IOF ОВ tan IOF cot IOF'= BH OH cox IOF= BH Thus, those ratios of an obtuse angle, which contain the base - OB (viz. all but the sine and cosecant), are negative. A fuller explanation of this adoption of the minus sign will be given later. But the student will immediately perceive its convenience, for by its means the formulæ connecting the ratios of the angles and the sides of a triangle are the same whether the triangle be acute-angled, right-angled or obtuseangled. 137. To find the ratios of the obtuse angles 180o – A and 90° + A in terms of those of the acute angle A. Take the figure of last article. -OB: cos IOF= (1) Let BOH = A. BH OH - ОВ cos IOF= =- cos BOH, i.e. cos (180° – A) =- cos A, OH and so on. (2) Let BHO = A. BH OH = cos BHO, i.e. sin (90° + A) = cos A, — ОВ – sin BHO, i.e. cos (90° + A)= - sin A, OH and so on. 138. To find the ratio of the acute angles 180° – A and Å – 90° in terms of those of the obtuse angle A. Take the same figure. BH ОН OB cos BOH= OH cos IOF, i.e. cos (180° – A) = – cos A, and so on. (4) Let IOF = A, then BHO= A – 90°, OB cos IOF, i.e. sin (A – 90°) = – cos A, sin IOF, i.e. cos (A – 90°) = sin A, and so on. 139. To find the ratio of 180°. cot 180° - cot 0 = cos BHO = = ve ve . Example. To trace the changes in the ratios of obtuse angles. sin A is + ve; and algebraically decreases from 1 to 0.. to 0. cox A is + ve; and algebraically increases from 1 to co. 140. From the proposition that the angles of a triangle are together equal to two right angles many important results follow. Thus, if A, B, C are the angles of a triangle, we have A + B + C= 180°, .. B+C=180° – A. [Art. 137. [Art. 137. = = COS Write down the ratios of 135°, 120°, 150°, 1121, 105o. 2. Show that sin 120° = 2 sin 60° cos 60° and that cos 120° = cosa 60° – sino 60°. = 3. Show that tan 150° = cot 60° cosec 60° and that cos? 135° = cos 60°. 4. If A, B, C are the angles of a triangle, show that sin } (B+C) = cos į A; and that tan 1 (A + B) = cot } C. 5. Show that the ratios of the half-angles of a triangle are all positive. 6. Show that, given the sine or the cosecant of an angle of a triangle, that angle may have either of two values. 7. Show that the area of a triangle is measured by half the product of two sides and the sine of the internal angle contained by them, whether that angle is acute, obtuse or right. |