Very Short Answer on Trigonometrical Functions Class 11 OP Malhotra Exe-4E ISC Maths Solutions Ch-4. In this article you would learn all type problem on Trigonometrical Function. Step by step solutions of latest textbook has been given as latest syllabus. Visit official Website CISCE for detail information about ISC Board Class-11 Mathematics.
Trigonometrical Functions Class 11 OP Malhotra Very Short Answer ISC Maths Solutions Ch-4
| Board | ISC |
| Publications | S Chand |
| Subject | Maths |
| Class | 11th |
| Chapter-4 | Trigonometrical Functions |
| Writer | OP Malhotra |
| Exe-4(E) | Very Short Answer Type Questions. |
Very Short Answer on Trigonometrical Functions
OP Malhotra ISC Class 11 Maths Solutions
Find the value of
Que-1: cos θ cos (90° − θ) − sin θ sin (90° − θ)
Sol: We know that :
cos(90° − θ) = sin θ
sin(90° − θ) = cos θ
So,
cos θ · sin θ − sin θ · cos θ
= 0
Que-2: sin 765°
Sol: sin 765° = sin (720° + 45°)
Using identity: sin (360°k + θ) = sin θ
sin 765° = sin 45°
= 1 / √2
= √2 / 2
Que-3: sin 120° cos 30° tan 135°
Sol: sin 120° = sin(180° − 60°)
= sin 60° = √3/2
cos 30° = √3/2
tan 135° = tan(180° − 45°)
= −tan 45° = −1
Therefore,
sin 120° × cos 30° × tan 135°
= (√3/2) × (√3/2) × (−1)
= (3/4) × (−1)
= −3/4
Que-4: [ cot 54° / tan 36° ] + [ tan 20° / cot 70° ]
Sol: Using identity : cot θ = tan (90° − θ)
cot 54° = tan 36°
cot 70° = tan 20°
Substitute in expression:
= (tan 36° / tan 36°) + (tan 20° / tan 20°)
= 1 + 1 = 2
Que-5: cos 255° + sin 165°
Sol: cos 255° = cos (180° + 75°) = −cos 75°
sin 165° = sin (180° − 15°) = sin 15°
So, cos 255° + sin 165° = −cos 75° + sin 15°
But, cos 75° = sin 15°
Therefore, −cos 75° + sin 15° = −sin 15° + sin 15° = 0
Que-6: cot (−15π/4)
Sol: cot(−15π/4)
Using periodicity of cot (period = π):
−15π/4 + 4π = π/4
So, cot(−15π/4) = cot(π/4)
cot(π/4) = 1
Que-7: cos 1° cos 2° … cos 100°
Sol: We know that:
cos(90° + θ) = −sinθ
In the given product, one of the terms is:
cos 90° = 0
Since the entire expression is a product and one factor is zero:
cos 1° × cos 2° × … × cos 100° = 0
Que-8: cos θ + sin θ if θ = 495°
Sol: Since 495° = 360° + 135°,
So, 495° is coterminal with 135°.
cos 495° = cos 135° = −1/√2
sin 495° = sin 135° = 1/√2
Therefore,
cosθ + sinθ = (−1/√2) + (1/√2) = 0
Que-9: cos (−1710°) sin 330°
Sol: cos(−1710°) = cos(1710°) [because cos(−θ) = cos θ]
1710° = 1710 − 1440 = 270°
⇒ cos(1710°) = cos(270°) = 0
sin(330°) = −1/2
∴ cos(−1710°) × sin(330°) = 0 × (−1/2)
= 0
Que-10: sec θ + tan θ if sin θ = 24/25 and θ lies in the second quadrant
Sol: Given: sin θ = 24/25, θ is in second quadrant
In second quadrant:
sin θ is positive and cos θ is negative
Using identity: sin²θ + cos²θ = 1
cos θ = √(1 − sin²θ)
= √(1 − (24/25)²)
= √(1 − 576/625)
= √(49/625)
= 7/25
Since θ is in 2nd quadrant ⇒ cos θ = -7/25
sec θ = 1/cos θ = -25/7
tan θ = sin θ / cos θ = (24/25)/(-7/25) = -24/7
Therefore,
sec θ + tan θ = (-25/7) + (-24/7)
= -49/7 = -7
Que-11: In ΔABC, if ∠A = π/2, then prove that cos²B + cos²C = 1.
Sol: In a right-angled triangle, the sum of the other two angles is:
B + C = π/2
So,
C = π/2 − B
Now,
cos C = cos(π/2 − B) = sin B
Therefore,
cos²C = sin²B
Now add:
cos²B + cos²C = cos²B + sin²B
= 1
Que-12: If tan x = −5/12, where x lies in the 2nd quadrant then cosec x = …………, sec x = …………
Sol: Given: tan x = −5/12 and x is in 2nd quadrant
In 2nd quadrant: sin x is positive, cos x is negative.
Let opposite = 5, adjacent = 12
Hypotenuse = √(5² + 12²) = 13
sin x = 5/13 cos x = −12/13
cosec x = 1/sin x = 13/5
sec x = 1/cos x = −13/12
Que-13: sin 120° cos 150° − cos 240° sin 330° = …………
Sol: sin 120° = √3/2, cos 150° = −√3/2
cos 240° = −1/2, sin 330° = −1/2
LHS = (√3/2)(−√3/2) − (−1/2)(−1/2)
= (−3/4) − (1/4)
= −4/4
= −1
Que-14: If sec θ and tan θ are the roots of ax² + bx + c = 0, then prove that sec θ − tan θ = −a/b.
Sol: Since secθ and tanθ are roots of the quadratic ax² + bx + c = 0,
Sum of roots:
secθ + tanθ = −b/a
Product of roots:
secθ · tanθ = c/a
Using identity:
sec²θ − tan²θ = 1
⇒ (secθ − tanθ)(secθ + tanθ) = 1
Substitute secθ + tanθ = −b/a:
(secθ − tanθ)(−b/a) = 1
⇒ secθ − tanθ = −a/b
Que-15: Prove that sin²(π/6) + cos²(π/3) − tan²(π/4) = −1/2.
Sol: We know standard values:
sin(π/6) = 1/2,
cos(π/3) = 1/2,
tan(π/4) = 1
Substitute values:
sin²(π/6) + cos²(π/3) − tan²(π/4)
= (1/2)² + (1/2)² − (1)²
= 1/4 + 1/4 − 1
= 1/2 − 1
= −1/2
Hence Proved.
Que-16: Prove that 2 sin²(π/6) + cosec²(7π/6) cos²(π/3) = −1/2.
Sol: sin(π/6) = 1/2 ⇒ sin²(π/6) = 1/4
cos(π/3) = 1/2 ⇒ cos²(π/3) = 1/4
sin(7π/6) = −1/2
⇒ cosec(7π/6) = −2
⇒ cosec²(7π/6) = 4
Now substitute:
LHS = 2(1/4) + 4(1/4)
= 1/2 + 1
= 3/2
Hence LHS ≠ RHS (−1/2)
Que-17: Prove that cot²(π/6) + cosec(5π/6) + 3 tan²(π/6) = 6.
Sol: We know the standard values:
cot(π/6) = √3 ⇒ cot²(π/6) = 3
sin(5π/6) = 1/2 ⇒ cosec(5π/6) = 2
tan(π/6) = 1/√3 ⇒ tan²(π/6) = 1/3
Now substitute:
cot²(π/6) + cosec(5π/6) + 3 tan²(π/6)
= 3 + 2 + 3(1/3)
= 3 + 2 + 1
= 6
Hence Proved
–: End Trigonometrical Functions Class 11 OP Malhotra Exe-4E ISC Maths Ch-4 Latest editions :–
Return to :- OP Malhotra ISC Class-11 S Chand Publication Maths Solutions
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