Question -
Answer -
(i)┬аtan 225o┬аcot 405o┬а+tan 765o┬аcot 675o┬а= 0
Let us consider LHS:
tan 225┬░ cot 405┬░ + tan 765┬░ cot 675┬░
tan (90┬░ ├Ч 2 + 45┬░) cot (90┬░ ├Ч 4 + 45┬░) + tan (90┬░ ├Ч 8+ 45┬░) cot (90┬░ ├Ч 7 + 45┬░)
We know that when n is odd, cot┬атЖТ┬аtan.
tan 45┬░ cot 45┬░ + tan 45┬░ [-tan 45┬░]
tan 45┬░ cot 45┬░ тАУ tan 45┬░ tan 45┬░
1 ├Ч 1 тАУ 1 ├Ч 1
1 тАУ 1
0 = RHS
тИ┤ LHS = RHS
Hence proved.
(ii)┬аsin 8╧А/3 cos 23╧А/6 + cos 13╧А/3 sin 35╧А/6 = 1/2
Let us consider LHS:
sin 8╧А/3 cos 23╧А/6 + cos 13╧А/3 sin 35╧А/6
sin 480┬░ cos 690┬░ + cos 780┬░ sin 1050┬░
sin (90┬░ ├Ч 5 + 30┬░) cos (90┬░ ├Ч 7 + 60┬░) + cos (90┬░ ├Ч 8+ 60┬░) sin (90┬░ ├Ч 11 + 60┬░)
We know that when n is odd, sin┬атЖТ┬аcos andcos┬атЖТ┬аsin.
cos 30┬░ sin 60┬░ + cos 60┬░ [-cos 60┬░]
тИЪ3/2 ├Ч тИЪ3/2 тАУ 1/2 ├Ч 1/2
3/4 тАУ 1/4
2/4
1/2
= RHS
тИ┤ LHS = RHS
Hence proved.
(iii)┬аcos 24o┬а+ cos 55o┬а+cos 125o┬а+ cos 204o┬а+ cos 300o┬а=1/2
Let us consider LHS:
cos 24o┬а+ cos 55o┬а+cos 125o┬а+ cos 204o┬а+ cos 300o
cos 24┬░ + cos (90┬░ ├Ч 1 тАУ 35┬░) + cos (90┬░ ├Ч 1 + 35┬░) +cos (90┬░ ├Ч 2 + 24┬░) + cos (90┬░ ├Ч 3 + 30┬░)
We know that when n is odd, cos┬атЖТ┬аsin.
cos 24┬░ + sin 35┬░ тАУ sin 35┬░ тАУ cos 24┬░ + sin 30┬░
0 + 0 + 1/2
1/2
= RHS
тИ┤ LHS = RHS
Hence proved.
(iv)┬аtan (-125o) cot (-405o) тАУ tan(-765o) cot (675o) = 0
Let us consider LHS:
tan (-125o) cot (-405o) тАУ tan(-765o) cot (675o)
We know that tan (-x) = -tan (x) and cot (-x) = -cot(x).
[-tan(225┬░)] [-cot (405┬░)] тАУ [-tan (765┬░)] cot (675┬░)
tan (225┬░) cot (405┬░) + tan (765┬░) cot (675┬░)
tan (90┬░ ├Ч 2 + 45┬░) cot (90┬░ ├Ч 4 + 45┬░) + tan (90┬░ ├Ч 8+ 45┬░) cot (90┬░ ├Ч 7 + 45┬░)
tan 45┬░ cot 45┬░ + tan 45┬░ [-tan 45┬░]
1 ├Ч 1 + 1 ├Ч (-1)
1 тАУ 1
0
= RHS
тИ┤ LHS = RHS
Hence proved.
(v)┬аcos 570o┬аsin 510o┬а+sin (-330o) cos (-390o) = 0
Let us consider LHS:
cos 570o┬аsin 510o┬а+sin (-330o) cos (-390o)
We know that sin (-x) = -sin (x) and cos (-x) = +cos(x).
cos 570o┬аsin 510o┬а+[-sin (330o)] cos (390o)
cos 570o┬аsin 510o┬атАУsin (330o) cos (390o)
cos (90┬░ ├Ч 6 + 30┬░) sin (90┬░ ├Ч 5 + 60┬░) тАУ sin (90┬░ ├Ч 3+ 60┬░) cos (90┬░ ├Ч 4 + 30┬░)
We know that cos is negative at 90┬░ + ╬╕ i.e. in Q2┬аandwhen n is odd, sin┬атЖТ┬аcos and cos┬атЖТ┬аsin.
-cos 30┬░ cos 60┬░ тАУ [-cos 60┬░] cos 30┬░
-cos 30┬░ cos 60┬░ + cos 60┬░ cos 30┬░
0
= RHS
тИ┤ LHS = RHS
Hence proved.
(vi)┬аtan 11╧А/3 тАУ 2 sin 4╧А/6 тАУ 3/4 cosec2┬а╧А/4+ 4 cos2┬а17╧А/6 = (3 тАУ 4тИЪ3)/2
Let us consider LHS:
tan 11╧А/3 тАУ 2 sin 4╧А/6 тАУ 3/4 cosec2┬а╧А/4+ 4 cos2┬а17╧А/6
tan (11 ├Ч 180o)/3 тАУ 2 sin (4 ├Ч 180o)/6тАУ 3/4 cosec2┬а180o/4 + 4 cos2┬а(17 ├Ч180o)/6
tan 660o┬атАУ 2 sin 120o┬атАУ3/4 (cosec 45o)2┬а+ 4 (cos 510o)2
tan (90┬░ ├Ч 7 + 30┬░) тАУ 2 sin (90┬░ ├Ч 1 + 30┬░) тАУ 3/4[cosec 45┬░]2┬а+ 4 [cos (90┬░ ├Ч 5 + 60┬░)]2
We know that tan and cos is negative at 90┬░ + ╬╕ i.e.in Q2┬аand when n is odd, tan┬атЖТ┬аcot,sin┬атЖТ┬аcos and cos┬атЖТ┬аsin.
[-cot 30┬░] тАУ2 cos 30┬░ тАУ 3/4 [cosec 45┬░]2┬а+ [-sin 60┬░]2
тАУ cot 30┬░ тАУ 2 cos 30┬░ тАУ 3/4 [cosec 45┬░]2┬а+[sin 60┬░]2
-тИЪ3 тАУ 2тИЪ3/2 тАУ 3/4 (тИЪ2)2┬а+ 4 (тИЪ3/2)2
-тИЪ3 тАУ тИЪ3 тАУ 6/4 + 12/4
(3 тАУ 4тИЪ3)/2
= RHS
тИ┤ LHS = RHS
Hence proved.
(vii)┬а3 sin ╧А/6 sec ╧А/3 тАУ 4 sin 5╧А/6 cot ╧А/4 = 1
Let us consider LHS:
3 sin ╧А/6 sec ╧А/3 тАУ 4 sin 5╧А/6 cot ╧А/4
3 sin 180o/6 sec 180o/3 тАУ 4 sin5(180o)/6 cot 180o/4
3 sin 30┬░ sec 60┬░ тАУ 4 sin 150┬░ cot 45┬░
3 sin 30┬░ sec 60┬░ тАУ 4 sin (90┬░ ├Ч 1 + 60┬░) cot 45┬░
We know that when n is odd, sin┬атЖТ┬аcos.
3 sin 30┬░ sec 60┬░ тАУ 4 cos 60┬░ cot 45┬░
3 (1/2) (2) тАУ 4 (1/2) (1)
3 тАУ 2
1
= RHS
тИ┤ LHS = RHS
Hence proved.