M 





(1). If tan - :. then Mihstituti due in (. 



Art. 14, given cot0 = |; rabetttutitig tfaeee Yataei in (7) ami (*) of the 

 same article gives the values of sec and of esc 6 ; and substituting those 

 values in (1) and (2) gives sin $ and cos 0. 



le. ABC with the sides 



AH - 1 ami / -n tBAC is an a. It 



I /; 1 and in ' :;. th.-n .1 C = VlU, and the other lune- 

 r lions ofti are at once seen to be : 



sin* 



8 



vlo* 



1 



vTo 



3 ' 



sec = V10, and cot d 



i ; ..:;. 



of tiii.se methods may be employed to solve the 

 other parts of this example; the second method is usually 

 to be preferred. 



4. By means of a right triangle, with appropriate acute 

 angles, find the numerical values of the trigonometric 

 ratios of the following angles : 



30; 45; 60; 90; 135; and -45. 



5. Express the following functions in terms of functions of 

 angles less than 90 : 



tan 8500; -esc 290; sin (-369 



6. Solve the following equations : 



-cosi^-; and cot(- 1215). 



(1) sin 6 = - cos 210; (2) cos0 = sin 20: 



COST 



sin x cot 2 x 

 and (4) (sec* - l)(cscx + 1)= |. 



7 In UK* following identities transform tin- fir-t member into the 

 second: 



tan - cot $ _ _' , . , AV sec x + esc x \ + rot r , 



1-cotz' 



0) 



(2) 



tan $ + cot csc 5 ' sec x - esc x 



(8) oscr(secz 1)- cotz(l -cosx) = tanx - sin a:; 

 (4) (2 r sin a cos a> -I- r*(cos*a - sin*a) f =r*; 



osacosA -f -in '/ -in A)* + (sinacos6 - r>*n >in/) 2 =l ; and 

 (6) (rcos^)* + (rsiu <cos0) 2 + (r.sin $ .tin ^) 2 =r. 



