18. (sec+cosec 8) (sin 8+cos 0) = sec @ cosec 8+2. 19. (cos - sin 0) (cosec - sec )=sec cosec ◊ – 2. 20. (1+cot+cosec ) (1+cot 0 - cosec 0) = 2 cot 0. 21. (sec 0+tan 0-1) (sec - tan 6+1)=2 tan 6. 22. (sin A+cosec A)2+(cos +sec A)2=tan2 A + cot2 A +7. 23. (sec2 4+tan2 A) (cosec2 A+cot2 A)=1+2 sec2 A cosec2 A. 24. (1-sin A+cos A)2=2 (1 − sin A) (1+cos A). 25. sin A (1+tan A)+cos A (1+cot A)=sec A+cosec A. 26. cos 0 (tan 0+2) (2 tan 6+1)=2 sec +5 sin 0. [The following examples contain functions of two angles; in each case the two angles are quite independent of each other.] 30. tan2 a+sec2 B=sec2 a+tan2 ß. 31. 33. cot a tan ẞß (tan a + cot ẞ)=cot a+tan B. 34. sin2 a cos2 ß – cos2 a sin2 ß=sin2 a - sin2 ß. 35. sec2 a tan2 B-tan2 a sec2 B=tan2 B- tan2 a. 36. (sin a cosẞ+cos a sin ẞ)2+(cos a cos B-sin a sin ß)2=1. 32. By means of the relations collected together in Art. 29, all the trigonometrical ratios can be expressed in terms of any one. Example 1. Express all the trigonometrical ratios of A in terms of tan A. OBS. In writing down the ratios we choose the simplest and most natural order. For instance, cot A is obtained at once by the reciprocal relation connecting the tangent and cotangent: sec A comes immediately from the tangent-secant formula; the remaining three ratios now readily follow. 33. It is always possible to describe a right-angled triangle when two sides are given: for the third side can be found by Euc. I. 47, and the construction can then be effected by Euc. I. 22. We can thus readily obtain all the trigonometrical ratios when one is given, or express all in terms of any one. 5 Example 1. Given cos A= find cosec A and cot A. 13' Take a right-angled triangle PQR, of which Q is the right angle, having the hypotenuse PR=13 units, and PQ-5 units. Example 2. Find tan A and cos A in terms of cosec A. Take a triangle PQR right-angled at Q, and having ▲ RPQ=A. For shortness, denote cosec A by c. Let QR be taken as the unit of measurement; then QR=1, and therefore PR=c. Let PQ contain x units; then R is 7. 5. Find the sine and cotangent of an angle whose secant 6. If 25 sin A=7, find tan A and sec A. 7. Express sin A and tan A in terms of cos A. Express cosec a and cos a in terms of cot a. 8. 9. Find sine and cot in terms of sec 0. 10. sin A. Express all the trigonometrical ratios of A in terms of 11. Given sin A-cos A=0, find cosec A. CHAPTER IV. TRIGONOMETRICAL RATIOS OF CERTAIN ANGLES. 34. Trigonometrical Ratios of 45°. Let BAC be a right-angled isosceles triangle, with the right angle at C; so that B=A=45°. Let each of the equal sides contain 7 units, The other three ratios are the reciprocals of these; thus cosec 45° = √2, sec 45° =√2, cot 45°=1; or they may be read off from the figure. |