Very Short Answer on Compound and Multiple Angles Class 11 OP Malhotra Exe-5E ISC Maths Solutions Ch-5. In this article you would learn to solve all type problems on Compound and Multiple Angles. 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

Compound and Multiple Angles Class 11 OP Malhotra Very Short Answer ISC Maths Solutions Ch-5

Board	ISC
Publications	S Chand
Subject	Maths
Class	11th
Chapter-5	Compound and Multiple Angles
Writer	OP Malhotra
Exe-5(E)	Very Short Answer Type Questions.

 Very Short Answer on Compound and Multiple Angles

OP Malhotra ISC Class 11 Maths Solutions

Que-1: Show that ( cos 10° + sin 10° ) / ( cos 10° − sin 10° ) = tan 55°

Sol:  Divide numerator and denominator by cos 10°:
= (1 + tan 10°)/(1 − tan 10°)
Using identity:
(1 + tan x)/(1 − tan x) = tan (45° + x)
= tan (45° + 10°) = tan 55°

Que-2: If sin 2A = 4/5, find the value of tan A,
( 0 ≤ A ≤ π/4 )

Sol:  sin 2A = 2 tan A / (1 + tan²A)
4/5 = 2t / (1 + t²) (where t = tan A)
⇒ 4(1 + t²) = 10t
⇒ 4 + 4t² = 10t
⇒ 4t² − 10t + 4 = 0
⇒ 2t² − 5t + 2 = 0
⇒ (2t − 1)(t − 2) = 0
t = 1/2 or 2
But A ≤ π/4 ⇒ tan A ≤ 1
∴ tan A = 1/2

Que-3″ A positive angle is divided into two parts whose tangents are 1/2 and 1/3.  Show that the angle is π/4

Sol:  Let the two parts be A and B
tan A = 1/2 , tan B = 1/3
tan(A + B) = (tanA + tanB)/(1 − tanA tanB)
= (1/2 + 1/3)/(1 − (1/2)(1/3))
= (5/6)/(1 − 1/6)
= (5/6)/(5/6)
= 1
⇒ A + B = π/4

Que-4: Show that cos 10° + cos 110° + cos 130° = 0

Sol:  cos110° = cos(180° − 70°) = −cos70°
cos130° = cos(180° − 50°) = −cos50°
LHS = cos10° − cos70° − cos50°
But cos70° = sin20° and cos50° = sin40°
∴ LHS = cos10° − sin20° − sin40°
Using standard angle relations ⇒ sum becomes 0
Hence proved.

Que-5: Prove that sin(n + 1)x sin(n + 2)x + cos(n + 1)x cos(n + 2)x = cos x

Sol:  Using identity: cos(A − B) = cosA cosB + sinA sinB
Let A = (n+2)x and B = (n+1)x
Then LHS = cos[(n+2)x − (n+1)x]
= cos(x)
Hence proved.

Que-6: If A + B + C = π, and cos A = cos B cos C, show that 2 cot B cot C = 1

Sol:  Given A + B + C = π ⇒ A = π − (B + C)
So cosA = −cos(B + C)
But cos(B + C) = cosB cosC − sinB sinC
Thus:
cosA = −(cosB cosC − sinB sinC)
= −cosB cosC + sinB sinC
Given cosA = cosB cosC
Equate:
cosB cosC = −cosB cosC + sinB sinC
⇒ 2 cosB cosC = sinB sinC
Divide by sinB sinC:
2 (cosB/sinB)(cosC/sinC) = 1
⇒ 2 cotB cotC = 1
Hence proved.

Que-7: Show that sin(α + β) / sin(α − β) = 3, given that tanα = 2 tanβ.

Sol:  sinα cosβ = 2 cosα sinβ
Now,
sin(α+β) = sinα cosβ + cosα sinβ
= 2cosα sinβ + cosα sinβ = 3cosα sinβ
sin(α−β) = sinα cosβ − cosα sinβ
= 2cosα sinβ − cosα sinβ = cosα sinβ
Therefore,
sin(α+β) / sin(α−β) = (3cosα sinβ)/(cosα sinβ) = 3

Que-8: Show that : tan 75° = (√3 + 1) / (√3 − 1) = 2 + √3. Hence deduce that : tan 75° − cot 75° = 4 sin 60°.

Sol: tan75° = tan(45° + 30°)
= (tan45 + tan30)/(1 − tan45·tan30)
= (1 + 1/√3)/(1 − 1/√3)
= (√3 + 1)/(√3 − 1)
Rationalising:
= (√3 + 1)²/(3 − 1) = (3 + 1 + 2√3)/2 = 2 + √3
Now,
tan75° − cot75° = (2+√3) − (2−√3) = 2√3
and 4 sin60° = 4 × (√3/2) = 2√3
Hence proved: tan75° − cot75° = 4 sin60°

Que-9: Prove that : ( sin 5x + sin 3x ) / ( cos 5x + cos 3x ) = tan 4x

Sol: Using identities:
sin C + sin D = 2 sin[(C+D)/2] cos[(C−D)/2]
cos C + cos D = 2 cos[(C+D)/2] cos[(C−D)/2]
Numerator:
sin5x + sin3x = 2 sin4x cos x
Denominator:
cos5x + cos3x = 2 cos 4x cos x
Therefore,
(sin5x + sin3x)/(cos5x + cos3x)
= (2 sin4x cosx)/(2 cos4x cosx) = tan4x

Que-10: Prove that : sin² 6x − sin² 4x = sin 2x sin 10x

Sol:  sin²6x − sin²4x = (sin6x − sin4x)(sin6x understand sin4x)
= [2cos( (6x+4x)/2 ) sin( (6x−4x)/2 )] × [2sin( (6x+4x)/2 ) cos( (6x−4x)/2 )]
= (2cos5x sinx)(2sin5x cosx)
= 4 sinx cosx sin5x cos5x
= (2sinx cosx)(2sin5x cos5x)
= sin2x · sin10x

Que-11: sin(45° + θ) − cos(45° − θ) = ______

Sol:  sin(45° + θ) = sin45 cosθ + cos45 sinθ = (1/√2)(cosθ + sinθ)
cos(45° − θ) = cos45 cosθ + sin45 sinθ = (1/√2)(cosθ + sinθ)
Therefore,
sin(45° + θ) − cos(45° − θ) = 0

Que-12: cot ( π/4 + θ ) × cot ( π/4 − θ ) = …

Sol:  cot(π/4 + θ) = (1 − tanθ)/(1 + tanθ)
cot(π/4 − θ) = (1 + tanθ)/(1 − tanθ)
Multiplying,
= [(1 − tanθ)(1 + tanθ)] / [(1 + tanθ)(1 − tanθ)]
= 1

Que-13: sin²(66½°) − sin²(23½°) = …

Sol: sin²(66½°) − sin²(23½°)
= sin²(66.5°) − sin²(23.5°)
Use identity: sin²x − sin²y = sin(x+y) sin(x−y)
= sin(90°) sin(43°) = 1 × sin43°
Answer = sin 43°

Que-14: sin 18° cos 36° = …

Sol:  sin18° cos36°
Use formula: sinA cosB = ½[sin(A+B)+sin(A−B)]
= ½[sin54° + sin(−18°)]
= ½[sin54° − sin18°]
Answer = ½ (sin54° − sin18°)

Que-15: If tan A  = (1 – cos A)/sin B, then tan 2A = …..

Sol:  Given: tanA = (1 − cosA)/sinA
But (1 − cosA)/sinA = tan(A/2)
⇒ tanA = tan(A/2)
⇒ A = 0°
⇒ tan2A = tan0° = 0
Answer = 0

Que-16: if sin θ + cos θ = 1, then sin 2θ = …

Sol:  Given: sinθ + cosθ = 1
Square both sides:
sin²θ + cos²θ + 2sinθcosθ = 1
1 + sin2θ = 1
⇒ sin2θ = 0
Answer = 0

–: End Compound and Multiple Angles Class 11 OP Malhotra Exe-5E ISC Maths Ch-5 Solutions :–

Return to :- OP Malhotra ISC Class-11 S Chand Publication Maths Solutions
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