Question

Prove that:
(i) tan 225^° cot 405^° + tan 765^° cot 675^° = 0

(ii) sin 8π/3 cos 23π/6 + cos 13π/3 sin 35π/6 = 1/2

(iii) cos 24^° + cos 55^° + cos 125^° + cos 204^° + cos 300^° = 1/2

(iv) tan (-125^°) cot (-405^°) – tan (-765^°) cot (675^°) = 0

(v) cos 570^° sin 510^° + sin (-330^°) cos (-390^°) = 0

(vi) tan 11π/3 – 2 sin 4π/6 – 3/4 cosec² π/4 + 4 cos² 17π/6 = (3 – 4√3)/2

(vii) 3 sin π/6 sec π/3 – 4 sin 5π/6 cot π/4 = 1

Answer 1

(i) Let us consider the LHS:

tan 225° cot 405° + tan 765° cot 675°

tan (90° × 2 + 45°) cot (90° × 4 + 45°) + tan (90° × 8 + 45°) cot (90° × 7 + 45°)

As we know that when n is odd, cot → tan.

tan 45° cot 45° + tan 45° [-tan 45°]

tan 45° cot 45° – tan 45° tan 45°

1 × 1 – 1 × 1

1 – 1

0 = RHS

∴ LHS = RHS

Thus proved.

(ii) sin 8π/3 cos 23π/6 + cos 13π/3 sin 35π/6 = 1/2

Let us consider the LHS:

sin 8π/3 cos 23π/6 + cos 13π/3 sin 35π/6

sin 480° cos 690° + cos 780° sin 1050°

sin (90° × 5 + 30°) cos (90° × 7 + 60°) + cos (90° × 8 + 60°) sin (90° × 11 + 60°)

As we know that when n is odd, sin → cos and cos → sin.

cos 30° sin 60° + cos 60° [-cos 60°]

√3/2 × √3/2 – 1/2 × 1/2

3/4 – 1/4

2/4

1/2

= RHS

∴ LHS = RHS

Thus proved.

(iii) Let us consider the LHS:

cos 24^° + cos 55^° + cos 125^° + cos 204^° + cos 300^°

cos 24° + cos (90° × 1 – 35°) + cos (90° × 1 + 35°) + cos (90° × 2 + 24°) + cos (90° × 3 + 30°)

As we know that when n is odd, cos → sin.

cos 24° + sin 35° – sin 35° – cos 24° + sin 30°

0 + 0 + 1/2

1/2

= RHS

∴ LHS = RHS

Thus proved.

(iv) Let us consider the LHS:

tan (-125^°) cot (-405^°) – tan (-765^°) cot (675^°)

As we know that tan (-x) = -tan (x) and cot (-x) = -cot (x).

[-tan (225°)] [-cot (405°)] – [-tan (765°)] cot (675°)

tan (225°) cot (405°) + tan (765°) cot (675°)

tan (90° × 2 + 45°) cot (90° × 4 + 45°) + tan (90° × 8 + 45°) cot (90° × 7 + 45°)

tan 45° cot 45° + tan 45° [-tan 45°]

1 × 1 + 1 × (-1)

1 – 1

0

= RHS

∴ LHS = RHS

Thus proved.

(v) Let us consider the LHS:

cos 570^° sin 510^° + sin (-330^°) cos (-390^°)

As we know that sin (-x) = -sin (x) and cos (-x) = +cos (x).

cos 570^° sin 510^° + [-sin (330^°)] cos (390^°)

cos 570^° sin 510^° – sin (330^°) cos (390^°)

cos (90° × 6 + 30°) sin (90° × 5 + 60°) – sin (90° × 3 + 60°) cos (90° × 4 + 30°)

As we know that cos is negative at 90° + θ i.e. in Q₂ and when n is odd, sin → cos and cos → sin.

-cos 30° cos 60° – [-cos 60°] cos 30°

-cos 30° cos 60° + cos 60° cos 30°

0

= RHS

∴ LHS = RHS

Thus proved.

(vi) tan 11π/3 – 2 sin 4π/6 – 3/4 cosec² π/4 + 4 cos² 17π/6 = (3 – 4√3)/2

Let us consider the LHS:

tan 11π/3 – 2 sin 4π/6 – 3/4 cosec² π/4 + 4 cos² 17π/6

tan (11 × 180^°)/3 – 2 sin (4 × 180^°)/6 – 3/4 cosec² 180^°/4 + 4 cos² (17 × 180^°)/6

tan 660^° – 2 sin 120^° – 3/4 (cosec 45^°)² + 4 (cos 510^°)²

tan (90° × 7 + 30°) – 2 sin (90° × 1 + 30°) – 3/4 [cosec 45°]² + 4 [cos (90° × 5 + 60°)]²

As we know that tan and cos is negative at 90° + θ i.e. in Q₂ and when n is odd, tan → cot, sin → cos and cos → sin.

[-cot 30°] – 2 cos 30° – 3/4 [cosec 45°]² + [-sin 60°]²

– cot 30° – 2 cos 30° – 3/4 [cosec 45°]² + [sin 60°]²

-√3 – 2√3/2 – 3/4 (√2)² + 4 (√3/2)²

-√3 – √3 – 6/4 + 12/4

(3 – 4√3)/2

= RHS

∴ LHS = RHS

Thus proved.

(vii) 3 sin π/6 sec π/3 – 4 sin 5π/6 cot π/4 = 1

Let us consider the LHS:

3 sin π/6 sec π/3 – 4 sin 5π/6 cot π/4

3 sin 180^°/6 sec 180^°/3 – 4 sin 5(180^°)/6 cot 180^°/4

3 sin 30° sec 60° – 4 sin 150° cot 45°

3 sin 30° sec 60° – 4 sin (90° × 1 + 60°) cot 45°

As we know that when n is odd, sin → cos.

3 sin 30° sec 60° – 4 cos 60° cot 45°

3 (1/2) (2) – 4 (1/2) (1)

3 – 2

1

= RHS

∴ LHS = RHS
Thus proved.