Page No 554:
Answer:
Page No 554:
Question 28:
Answer:
From the given right-angled triangle, we have:
Page No 554:
Question 29:
From the given right-angled triangle, we have:
Answer:
From the given right-angled triangle, we have:
Page No 554:
Question 30:
From the given right-angled triangle, we have:
Answer:
From right-angled âABC, we have:
â
Page No 572:
Question 1:
From right-angled âABC, we have:
â
Answer:
On substituting the values of various T-ratios, we get:
sin 60o cos 30o + cos 60o sin 30o
Page No 572:
Question 2:
On substituting the values of various T-ratios, we get:
sin 60o cos 30o + cos 60o sin 30o
Page No 572:
Answer:
On substituting the values of various T-ratios, we get:
cos 60o cos 30o − sin 60o sin 30o
Page No 572:
Question 4:
On substituting the values of various T-ratios, we get:
cos 60o cos 30o − sin 60o sin 30o
Answer:
On substituting the values of various T-ratios, we get:
cos 45o cos 30o + sin 45o sin 30o
Page No 572:
Question 5:
On substituting the values of various T-ratios, we get:
cos 45o cos 30o + sin 45o sin 30o
Answer:
As we know that,
Page No 572:
Question 6:
As we know that,
Answer:
On substituting the values of various T-ratios, we get:
2 cos2 60o + 3 sin2 45o − 3 sin2 30o + 2 cos2 90o
Page No 572:
Question 7:
On substituting the values of various T-ratios, we get:
2 cos2 60o + 3 sin2 45o − 3 sin2 30o + 2 cos2 90o
Answer:
As we know that,
â
Page No 572:
Question 8:
As we know that,
â
Answer:
On substituting the values of various T-ratios, we get:
cot2 30o − 2 cos2 30o − sec2 45o + cosec2 30o
Page No 572:
Question 9:
On substituting the values of various T-ratios, we get:
cot2 30o − 2 cos2 30o − sec2 45o + cosec2 30o
Answer:
As we know that,
â
Page No 572:
Question 10:
As we know that,
â
Answer:
(i)
Hence, LHS = RHS
∴
(ii)
Hence, LHS = RHS
∴ 1−sin60°cos60°
Page No 572:
Question 11:
(i)
Hence, LHS = RHS
∴
(ii)
Hence, LHS = RHS
∴ 1−sin60°cos60°
Answer:
(i) sin 60o cos 30o − cos 60o sin 30o
∴ sin 60o cos 30o − cos 60o sin 30o = sin 30o
(ii) cos 60o cos 30o + sin 60o sin 30o
â
∴ cos 60o cos 30o + sin 60o sin 30o = cos 30o
(iii) 2 sin 30o cos 30o
∴ 2 sin 30o cos 30o = sin 60o
(iv) 2 sin 45o cos 45o
Also, sin 90o = 1
∴ 2 sin 45o cos 45o = sin 90o
Page No 572:
Question 12:
(i) sin 60o cos 30o − cos 60o sin 30o
∴ sin 60o cos 30o − cos 60o sin 30o = sin 30o
(ii) cos 60o cos 30o + sin 60o sin 30o
â
∴ cos 60o cos 30o + sin 60o sin 30o = cos 30o
(iii) 2 sin 30o cos 30o
∴ 2 sin 30o cos 30o = sin 60o
(iv) 2 sin 45o cos 45o
Also, sin 90o = 1
∴ 2 sin 45o cos 45o = sin 90o
Answer:
A = 45
o
⇒ 2
A = 2
45
o = 90
o
(i) sin 2
A = sin 90
o = 1
2 sin
A cos
A = 2 sin 45
o cos 45
o =
∴ sin 2
A = 2 sin
A cos
A
(ii) cos 2
A = cos 90
o = 0
2 cos
2A − 1 = 2 cos
2 45
o − 1 =
Now, 1 − 2 sin
2A =
∴ cos 2
A =
2 cos
2A − 1 = 1 − 2 sin
2A
Page No 572:
Question 13:
A = 45
o
⇒ 2
A = 2
45
o = 90
o
(i) sin 2
A = sin 90
o = 1
2 sin
A cos
A = 2 sin 45
o cos 45
o =
∴ sin 2
A = 2 sin
A cos
A
(ii) cos 2
A = cos 90
o = 0
2 cos
2A − 1 = 2 cos
2 45
o − 1 =
Now, 1 − 2 sin
2A =
∴ cos 2
A =
2 cos
2A − 1 = 1 − 2 sin
2A
Answer:
A = 30
o
⇒ 2
A = 2
30
o = 60
o
â(i) sin 2A = sin 60
o =
∴
(ii) cos 2A = cos 60
o =
∴
(iii) tan 2A = tan 60o =
∴ =2tanA1+tan2A
Page No 572:
Question 14:
A = 30
o
⇒ 2
A = 2
30
o = 60
o
â(i) sin 2A = sin 60
o =
∴
(ii) cos 2A = cos 60
o =
∴
(iii) tan 2A = tan 60o =
∴ =2tanA1+tan2A
Answer:
A = 60
o and
B = 30
o
Now,
A +
B = 60
o + 30
oâ = 90
o
Also,
A −
B = 60
o − 30
o = 30
o
(i) sin (
A +
B) = sin 90
o = 1
sin
A cos
B + cos
A sin
B = sin 60
o cos 30
o + cos 60
o sin 30
o
=
∴ sin (
A +
B) = sin
A cos
B + cos
A sin
B
(ii) cos (
A +
B) = cos 90
o = 0
cos
A cos
B − sin
A sin
B =cos 60
o cos 30
o − sin 60
o sin 30
o
∴â cos (
A +
B) = cos
A cos
B − sin
A sin
B
Page No 573:
Question 15:
A = 60
o and
B = 30
o
Now,
A +
B = 60
o + 30
oâ = 90
o
Also,
A −
B = 60
o − 30
o = 30
o
(i) sin (
A +
B) = sin 90
o = 1
sin
A cos
B + cos
A sin
B = sin 60
o cos 30
o + cos 60
o sin 30
o
=
∴ sin (
A +
B) = sin
A cos
B + cos
A sin
B
(ii) cos (
A +
B) = cos 90
o = 0
cos
A cos
B − sin
A sin
B =cos 60
o cos 30
o − sin 60
o sin 30
o
∴â cos (
A +
B) = cos
A cos
B − sin
A sin
B
Answer:
(i) sin (A − B) = sin 30o =
sin A cos B − cos A sin B = sin 60o cos 30o − cos 60o sin 30o
=
∴ sin (A − B) = sin A cos B − cos A sin B
(ii) cos (A − B) = cos 30o =
cos A cos B + sin A sin B = cos 60o cos 30o + sin 60o sin 30o
=
∴â cos (A − B) = cos A cos B + sin A sin B
(iii) tan (A − B) = tan 30o =
∴â tan (A − B) =
Page No 573:
Question 16:
(i) sin (A − B) = sin 30o =
sin A cos B − cos A sin B = sin 60o cos 30o − cos 60o sin 30o
=
∴ sin (A − B) = sin A cos B − cos A sin B
(ii) cos (A − B) = cos 30o =
cos A cos B + sin A sin B = cos 60o cos 30o + sin 60o sin 30o
=
∴â cos (A − B) = cos A cos B + sin A sin B
(iii) tan (A − B) = tan 30o =
∴â tan (A − B) =
Answer:
Given:
13
Page No 573:
Question 17:
Given:
13
Answer:
A = 30o
⇒ 2A = 2 30o = 60o
By substituting the value of the given T-ratio, we get:
∴ tan 60o =
Page No 573:
Question 18:
A = 30o
⇒ 2A = 2 30o = 60o
By substituting the value of the given T-ratio, we get:
∴ tan 60o =
Answer:
A = 30
o
⇒ 2
A = 2
30
o = 60
o
By substituting the value of the given T-ratio, we get:
∴ cos 30
o =
Page No 573:
Question 19:
A = 30
o
⇒ 2
A = 2
30
o = 60
o
By substituting the value of the given T-ratio, we get:
∴ cos 30
o =
Answer:
A = 30o
⇒ 2A = 2 30o = 60o
By substituting the value of the given T-ratio, we get:
∴ sin 30o =
Page No 573:
Question 20:
A = 30o
⇒ 2A = 2 30o = 60o
By substituting the value of the given T-ratio, we get:
∴ sin 30o =
Answer:
Disclaimer:
can also be calculated by taking
.
Page No 573:
Question 21:
Disclaimer:
can also be calculated by taking
.
Answer:
As we know that,
â
Page No 573:
Question 22:
As we know that,
â
Answer:
Page No 573:
Question 23:
Answer:
As we know that,
Page No 573:
Question 24:
As we know that,
Answer:
As we know that,
Page No 573:
Question 25:
As we know that,
Answer:
Here, tan (A − B) =
⇒ tan (A − B) = tan 30o [âµ tan 30o = ]
⇒ A − B = 30o ...(i)
Also, tan (A + B) =
⇒â tan (A + B) = tan 60o [âµ tan 60o = ]
⇒ A + B = 60o ...(ii)
Solving (i) and (ii), we get:
A = 45o and B = 15o
Page No 574:
Question 26:
Here, tan (A − B) =
⇒ tan (A − B) = tan 30o [âµ tan 30o = ]
⇒ A − B = 30o ...(i)
Also, tan (A + B) =
⇒â tan (A + B) = tan 60o [âµ tan 60o = ]
⇒ A + B = 60o ...(ii)
Solving (i) and (ii), we get:
A = 45o and B = 15o
Answer:
As we know that,
Page No 574:
Question 31:
As we know that,
Answer:
(i) As we know that,
â
(ii) As we know that,
â
Page No 576:
Question 1:
(i) As we know that,
â
(ii) As we know that,
â
Answer:
As we know that,
Page No 576:
Question 2:
As we know that,
Answer:
As we know that,
Page No 576:
Question 3:
As we know that,
Answer:
As we know that,
Page No 576:
Question 4:
As we know that,
Answer:
As we know that,
Page No 576:
Question 5:
As we know that,
Answer:
As we know that,
â
Page No 576:
Question 6:
As we know that,
â
Answer:
As we know that,
â
Page No 576:
Question 7:
As we know that,
â
Answer:
As we know that,
â
Page No 576:
Question 8:
As we know that,
â
Answer:
Given: tanx = 3cotx
â
Page No 576:
Question 9:
Given: tanx = 3cotx
â
Answer:
As we know that,
â
Page No 576:
Question 10:
As we know that,
â
Answer:
As we know that,
â
Page No 576:
Question 11:
As we know that,
â
Answer:
As we know that,
â
Page No 576:
Question 12:
As we know that,
â
Answer:
As we know that,
â
Page No 577:
Question 13:
As we know that,
â
Answer:
As we know that,
â
Page No 577:
Question 14:
As we know that,
â
Answer:
â
Page No 577:
Question 15:
â
Answer:
As we know that,
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