(36) tan (45° + A) – tan (45o – A)=2 tan 24.

(37) tan (45°-4)+cot (45o – A) = 2 sec 24.

NOTE. The similarity of these two results is apt to cause confusion. This may be avoided by observing that the second formula must be true when A=0°; and then cos 34=cos 0o=1. In which case the formula gives cos 0°=4 cos 0o – 3 cos 0o, or 1=4-3, which is true.

The first formula may be proved thus:

sin 34=sin (24+4)=sin 24. cos A+cos 24. sin A

=

(2 sin A. cos 4) cos A+ (1 - 2 sin2 A) sin A

=2 sin A. cos2 A+ sin A-2 sin3 A

=2 sin A (1 - sin2 A) + sin A − 2 sin3 A

=2 sin A-2 sin3 A+ sin A - 2 sin3 A

=3 sin A-4 sin3 A.

The second formula may be proved in a similar manner.

(3) sin (n+1)a.cos (n − 1)a - sin 2a=sin (n-1) a. cos (n + 1)a.

(6) If A=18o, prove that sin 2A=cos 34 ; hence prove that

√5-1
4

sin a + sin ẞ+ sin (a +ẞ)
sin a+ sin 3-sin (a+8)=cot. cot

(8) sin 24. sin 2B=sin2 (A + B) – sin2 (A – B). (9) cos 44-8 cos4 A-8 cos2 A+ 1.

(10) tan 500+ cot 50o=2 sec 10o.

(11) sin 34=4 sin A. sin (60° + A) sin (60° – A).

2

(12) (cot 4-tan 4)
4)* (cot 4 – 2 cot 24) = 4 cot A.

2

sin 3a + sinẞ. cos 5a - sin 7a

(13)

cos 3a-sinẞ.sin 5a-cos 7a

is independent of ß.

(14) (cosx+cos y)2 + (sin x + sin y)2=4 cos2

(15) 2 cos2 A. cos2 B+2 sin2 A. sin2 B=1+ cos 24. cos 2B.