30. sin1 a-cost a=2 sin2 a-1-1-2 cos2 a.

31. sec1 a-1=2 tan2 a+tan1 a.

32. cosec1 a-1=2 cot2 a+cot1 a.

33. (tan a cosec a)2 — (sin a sec a)2 = 1.

34. (sec o cot 0)2 - (cos e cosec )2 = 1. 35. tan20-cot20=sec2 0 - cosec2 0.

31. The foregoing examples have required little more than a direct application of the fundamental formula; we shall now give some identities offering a greater variety of treatment.

Example 1. Prove that sec2 4+cosec2 A = sec2 A cosec2 A.

Occasionally it is found convenient to prove the equality of the two expressions by reducing each to the same form.

Example 2. Prove that

sin2 A tan A+ cos2 A cot A+ 2 sin A cos Atan A+ cot A.

The transformations in the successive steps are usually suggested by the form into which we wish to bring the result. For instance, in this last example we might have proved the identity by substituting for the tangent and cotangent in terms of the sine and cosine. This however is not the best method, for the form in which the right-hand side is given suggests that we should retain tan a and cot ẞ unchanged throughout the work.

6. tan+cot = sec e cosec 0.

7. √1+cot2 A.√/sec2 A-1.√1-— sin2 A = 1.

8. (cos + sin 6)2+(cos - sin 0)2=2.

9. (1+tan 0)2 + (1 - tan 0)2 = 2 sec2 0.

10. (cot 0-1)2+(cot + 1)2=2 cosec2 0.

11. sin2 A (1+cot2 A) + cos2 A (1 + tan2 A) = 2.

12. cos2 A (sec2 A − tan2 A)+sin2 A (cosec2 A - cot2 A)=1.

13. cot2 a+cot1 a= cosec1 a - cosec2 a.

18. (sec +cosec ) (sin

+ cos 0) = sec cosec +2. 19. (cos 0-sin ) (cosec - sec )=sec e cosec - 2. 20. (1+cote+cosec ) (1+cot - cosec )=2 cot 0.

21. (sec 0+tan 0-1) (sec - tan 6+1)=2 tan 0.

22. (sin A+cosec A)2+(cos A+ sec A)2=tan2 A + cot2 A +7. 23. (sec2 A+tan2 A) (cosec2 A+cot2 4)=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 4)=sec A+cosec A.

26. cos 0 (tan 0+2) (2 tan 6+1)=2 sec 0+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 B.

33. cot a tan ẞ (tan a + cot B)=cot a+tan B.

34.

sin2 a cos2 B-cos2 a sin2 ß=sin2 a - sin2 ß.

35.

sec2 a tan2 ß-- tan2 a sec2 B=tan2 ß- tan2 a.

36.

(sin a cos B+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. 1. 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.

Now

cos RPQ=PR

so that

5

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.