Given that A = 30°

Then, LHS = sin3A

= sin3 × 30°

= sin90°

= 1

RHS = 3sinA - 4sin3A

= 3sin30° - 4(sin30°)3

= 3$$\frac{1}{2}$$ - 4($$\frac{1}{2}$$)3

= $$\frac{3}{2}$$ - 4 $$\frac{1}{8}$$

= $$\frac{3}{2}$$ - $$\frac{1}{2}$$

= $$\frac{2}{2}$$

= 1

$$\therefore$$ LHS = RHS verified.

Solution

LHS = cos 60

= $$\frac{1}{2}$$

RHS = cos230° - sin230°

($$\frac{\sqrt(3)}{2}$$)2 - ($$\frac{1}{2}$$)2

$$\frac{3}{4}$$ - $$\frac{1}{4}$$

$$\frac{2}{4}$$

$$\frac{1}{2}$$

$$\therefore$$ LHS = RHS proved.

Solution

sec2$$\frac{π}{4}$$ sec2$$\frac{π}{3}$$(cosec$$\frac{π}{6}$$ - cosec$$\frac{π}{2}$$)

= sec2 $$\frac{180°}{4}$$ × sec2$$\frac{180°}{3}$$ (cosec$$\frac{180°}{6}$$ - cosec$$\frac{180°}{2}$$)

= sec245° × sec260° (cosec30° - cosec90°)

= ($$\sqrt(2)$$)2 × (2)2 (2-1)

= 2×4×1

= 8

soln: L.H.S = sin4200 cos 3900 + cos (-3000) sin (-3300)

= sin4200 cos3900 + cos3000 (-sin3300)

= sin4200 cos3900 - cos3000 sin3300

= sin (3600 + 600) cos (3600 + 300) - cos (3600-60o) sin(3600-300)

= sin600 cos300 cos600 (-sin 300) = sin600 cos300 + cos600sin300

= $$\frac{(\sqrt{3})}{2}$$×$$\frac{(\sqrt{3})}{2}$$+$$\frac{1}{2}$$×$$\frac{1}{2}$$=$$\frac{3}{4}$$+$$\frac{1}{4}$$=$$\frac{3+1}{4}$$=$$\frac{4}{4}$$=1= R.H.S. proved.

Solution

Here, In ABC

B = 90° (Since Ac is the hypotenuse.)

So, using Pythagoras theorem

(AC)2 = (AB)2 + (BC)2

= (1)2 +($$\sqrt(3)$$)2

= 1 + 3 = 4
or, AC = 2 = b

Now, sinA =$$\frac{BC}{AC}$$

or, sinA = $$\frac{\sqrt(3)}{2}$$

or, sinA = sin 60°

$$\therefore$$ A = 60°

Again,

$$\angle$$A + $$\angle$$B + $$\angle$$C = 180°

or, 60° + 90° + $$\angle$$C = 180°

C = 180° - 150° = 30°

Hence, b = 2, A = 60° and C = 30°