Prove that:
Question 1: sin2 +cos2 – tan2 =
Solution:
Taking LHS in consideration, we get
= sin2
+cos2 – tan2 Substituting the values,
=
=
– 1 =
Hence, LHS = RHS
Question 2: 2sin2 +cosec2 cos2 =
Solution:
Taking LHS in consideration, we get
= 2sin2
+cosec2 cos2 = 2sin2
+ (- cosec cos2 Substituting the values,
=
= 2
+ 4 =
+ 1 =
Hence, LHS = RHS
Question 3: cot2 + cosec + 3tan2 = 6
Solution:
Taking LHS in consideration, we get
= cot2
+ cosec + 3tan2 = cot2
+ cosec + 3tan2 Substituting the values,
= (√3)2 + 2 + 3
= 3 + 2 + 3
= 6
Hence, LHS = RHS
Question 4: 2sin2 + 2cos2 + 2sec2 = 10
Solution:
Taking LHS in consideration, we get
= 2sin2
+ 2cos2 + 2sec2 = 2sin2
+ 2cos2 + 2sec2 Substituting the values,
=
= 2
+ 2(4) = 1 + 1 + 8
= 10
Hence, LHS = RHS
Question 5: Find the value of:
(i) sin 75°
Solution:
As, we don’t know the angle value for 75°, so we will break into the angles which we know.
75° = 30° + 45°, so lets use this and solve for sin(30° + 45°)
Using the trigonometric formula,
sin (A+B) = sin A cos B + cos A sin B
sin(30° + 45°) = sin 30° cos 45° + cos 30° sin 45°
Substituting values, we get
sin(75°) =
sin(75°) =
sin(75°) =
(ii) tan 15°
Solution:
As, we don’t know the angle value for 15°, so we will break into the angles which we know.
15° = 60° – 45° or 45° – 30° so lets use this and solve for tan(45° – 30°)
Using the trigonometric formula,
tan (A-B) =
tan(45° – 30°) =
Substituting values, we get
tan(15°) =
tan(15°) =
tan(15°) =
Now rationalizing the denominator, multiply and divide by
tan(15°) =
tan(15°) =
tan(15°) =
tan(15°) =
tan(15°) =
tan(15°) = 2 –
Prove the following:
Question 6: = sin (x+y)
Solution:
Taking LHS in consideration, we get
=
As, here there is multiplication of cos cos and sin sin, we will use Defactorisation Formulae,
2 cos A cos B = cos (A+B) + cos (A-B) and, ……………….(1)
2 sin A sin B = cos (A-B) – cos (A+B) ……………….(2)
Multiply and divide LHS by 2, we get
=
=
Subtracting (2) from (1) and using identity formulae, we get
2 cos A cos B – 2 sin A sin B = cos (A+B) + cos (A-B) – (cos (A-B) – cos (A+B))
2 cos A cos B – 2 sin A sin B = 2 cos (A+B)
Hence, using this
=
= cos
= cos
= sin (x+y) (As cos
= sin θ) Hence, LHS = RHS
Question 7:
Solution:
Taking LHS in consideration, we get
As, using Factorisation Formulae of tan, we have
tan (A+B) =
and, tan (A-B) =
Now, substituting the values
=
=
=
=
Hence, LHS = RHS
Question 8: = cot2x
Solution:
Taking LHS in consideration, we get
=
As, we know these standard values
cos(-x) = cos x
cos
= – cos x sin
= sin x cos
= – sin x Substituting these values, we have
=
=
= cot2 x
Hence, LHS = RHS
Question 9: = 1
Solution:
Taking LHS in consideration, we get
=
As, we know these standard values
cos
= sin x cos
= cos x cot
= cot x cot
= tan x Substituting the values, we have
=
=
=
As sin2 x + cos2 x = 1
= 1
Hence, LHS = RHS
Question 10: sin(n + 1)x sin(n + 2)x + cos(n + 1)x cos(n + 2)x = cos x
Solution:
Taking LHS in consideration, we get
= sin(n + 1)x sin(n + 2)x + cos(n + 1)x cos(n + 2)x
As, here there is multiplication of cos cos and sin sin, we will use Defactorisation Formulae,
2 cos A cos B = cos (A+B) + cos (A-B) and, ……………….(1)
2 sin A sin B = cos (A-B) – cos (A+B) ……………….(2)
Multiply and divide LHS by 2, we get
=
(sin(n + 1)x sin(n + 2)x + cos(n + 1)x cos(n + 2)x) =
(2 sin(n + 1)x sin(n + 2)x + 2 cos(n + 1)x cos(n + 2)x) Adding (1) and (2) and using identity formulae, we get
2 cos A cos B + 2 sin A sin B = cos (A+B) + cos (A-B) + cos (A-B) – cos (A+B)
2 cos A cos B + 2 sin A sin B = 2 cos (A-B)
Hence, using this
=
(2 cos((n + 1)x – (n + 2)x)) = cos((n + 1)x – (n + 2)x)
= cos (x-2x)
= cos (- x)
= cos x (As, cos(-x) = cos x)
Hence, LHS = RHS
Question 11: = – sin x
Solution:
Taking LHS in consideration, we get
=
Using the identity,
cos A – cos B = 2 sin
sin Substituting the values,
=
= 2 sin
sin = 2 sin
sin (-x) = 2 sin
sin (-x) = 2 ( sin
) sin (-x) = 2
(- sin x) =
Rationalizing the denominator, by multiplying and dividing by
=
=
=
=
sin x Hence, LHS = RHS
Question 12: sin2 6x – sin2 4x = sin 2x sin 10x
Solution:
Taking LHS in consideration, we get
= sin2 6x – sin2 4x
= sin 6x sin 6x – sin 4x sin 4x
As, here there is multiplication of sin sin, we will use Defactorisation Formulae,
2 sin A sin B = cos (A-B) – cos (A+B)
Multiply and divide LHS by 2, we get
=
(sin 6x sin 6x – sin 4x sin 4x) =
(2 sin 6x sin 6x – 2 sin 4x sin 4x) Using the identity, we can simplify
=
[(cos(6x-6x) – cos(6x+6x)) – (cos(4x-4x) – cos(4x+4x))] =
[(cos(0) – cos(12x)) – (cos(0) – cos(8x))] =
[1 – cos(12x) – 1 + cos(8x)] (As, cos 0 = 1) =
[cos(8x) – cos (12x)] Now, using the identity
cos A – cos B = 2 sin
sin Substituting the values, we have
=
= sin
sin = sin (10 x) sin (2x)
Hence, LHS = RHS
Question 13: cos2 2x – cos2 6x = sin 4x sin 8x
Solution:
Taking LHS in consideration, we get
= cos2 2x – cos2 6x
= cos 2x cos 2x – cos 6x cos 6x
As, here there is multiplication of cos cos, we will use Defactorisation Formulae,
2 cos A cos B = cos (A+B) + cos (A-B)
Multiply and divide LHS by 2, we get
=
(cos 2x cos 2x – cos 6x cos 6x) =
(2 cos 2x cos 2x – 2 cos 6x cos 6x) Using the identity, we can simplify
=
[(cos(2x+2x) + cos(2x-2x)) – (cos(6x+6x) + cos(6x-6x))] =
[(cos(2x+2x) + cos(0)) – (cos(6x+6x) + cos(0))] =
[(cos(4x) + 1 – cos(12x) – 1)] (As, cos 0 = 1) =
[cos(4x) – cos(12x)] Now, using the identity
cos A – cos B = 2 sin
sin Substituting the values, we have
=
= sin
sin = sin (8x) sin (4x)
Hence, LHS = RHS
Question 14: sin 2x + 2 sin 4x + sin 6x = 4 cos2 x sin 4x
Solution:
Taking LHS in consideration, we get
sin 2x + 2 sin 4x + sin 6x
After rearranging, we have
= (sin 2x + sin 6x) + 2 sin 4x
Using the identity, we can simplify
sin A+ sin B = 2 sin
cos = 2 sin
cos + 2 sin 4x = 2 sin
cos + 2 sin 4x = 2 sin (4x) cos (2x) + 2 sin 4x
Taking (2 sin 4x), we have
= 2 sin (4x) (cos (2x) + 1)
= 2 sin (4x) (2 cos2 x – 1 + 1) (As, cos 2θ = 2cos2 θ – 1)
= 2 sin (4x) (2 cos2 x)
= 4 sin (4x) cos2 x
Hence, LHS = RHS