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°