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Conditional Identities Involving The Angles Of A Triangle

Class 11th Mathematics RS Aggarwal Solution
Exercise 16
  1. sin 2A + sin 2B – sin 2C = 4cos A cos B sin C If A + B + C = π, prove that…
  2. cos 2A – cos 2B – cos 2C = -1 + 4 cos A sin B sin C If A + B + C = π, prove…
  3. cos 2A – cos 2B + cos 2C = 1 – 4sin A cos B sin C If A + B + C = π, prove that…
  4. sina+sinb+sinc = 4cos {a}/{2} cos frac {b}/{2} cos frac {c}/{2} If A + B + C…
  5. cosa+cosb+cosc = 1+4sin {a}/{2} sin frac {b}/{2} sin frac {c}/{2} If A + B +…
  6. {sin2a+sin2b+sin2c}/{sina+sinb+sinb+sinc} = 8sin frac {a}/{2} sin frac {b}/{2}…
  7. sin (B + C – A) + sin (C + A – B) – sin (A + B – C) = 4cos A cos B sin C If A +…
  8. {cosa}/{sinbsinc} + frac {cosb}/{sincsina} + frac {cosc}/{sinasinb} = 2 If A +…
  9. cos2 A + cos2 B + cos2 C = 1 – 2cos A cos B cos C If A + B + C = π, prove that…
  10. sin2 A – sin2 B + sin2 C = 2sin A cos B sin C If A + B + C = π, prove that…
  11. sin^{2} {a}/{2} + sin^{2} frac {b}/{2} + sin^{2} frac {c}/{2} = 1-2sin frac…
  12. tan 2A + tan 2B + tan 2C = tan 2A tan 2B tan 2C If A + B + C = π, prove that…
Exercise 16
Question 1.

If A + B + C = π, prove that

sin 2A + sin 2B – sin 2C = 4cos A cos B sin C
Answer:

= sin 2A + sin 2B – sin 2C


= 2 sin ( B + C ) cos A + 2 sin ( A + C ) cos B - 2 sin ( A + B ) cos C
using formula,
sin (A + B) = sin A cos B + cos A sin B
= sin 2A + sin 2B - sin 2C


Using formula


sin2A = 2sinAcosA
= 2sinAcosA + 2sinBcosB - 2sinCcosC


since A + B + C = π



And sin(π – A) = sinA
= 2sin(B + C)cos A + 2sin(A + C)cosB - 2sin(A + B)cosC
= 2 ( sin B cos C + cos B sin C ) cos A + 2(sinAcosC + cosAsinC)cosB - 2(sinAcosB + cosAsinB )cosC
= 2cosAsinBcosC + 2cosAcosBsinC + 2sinAcosBcosC + 2cosAcosBsinC– 2sinAcosBcosC – 2cosAsinBcosC
= 2cosAcosBsinC + 2cosAcosBsinC
= 4cosAcosBsinC


= R.H.S
Question 2.

If A + B + C = π, prove that

cos 2A – cos 2B – cos 2C = -1 + 4 cos A sin B sin C
Answer:

= cos2A – (cos2B + cos2C)


Using formula




= cos2A - {2cos(B+C)cos(B-C)}


since A + B + C = π



= cos2A – {2cos(π – A)cos(B-C)}


And cos(π – A) = -cosA


= cos2A – {-2cosAcos(B-C)}


= cos2A + 2cosAcos(B-C)


Using cos2A = 2cos2A -1


= 2cos2A – 1 + 2cosAcos(B-C)


= 2cosA{cosA + cos(B-C)} – 1







= 2cosA{2sinCsinB} – 1


= 4cosAsinBsinC – 1


= R.H.S
Question 3.

If A + B + C = π, prove that

cos 2A – cos 2B + cos 2C = 1 – 4sin A cos B sin C
Answer:

= cos2A – cos2B + cos2C


Using,




= cos2A - {2sin(B+C)sin(B-C)}


since A + B + C = π



And sin(π – A) = sinA


= cos2A – {2sin(π – A)sin(B-C)}


= cos2A – {2sinAsin(B-C)}


= cos2A - 2sinAsin(B-C)


Using , cos2A = 1 – 2sin2A


= -2sin2A + 1 – 2sinAsin(B-C)


= -2sinA{sinA + sin(B-C)} + 1







= -2sinA{2cosCsinB} + 1


= -4sinAcosBsinC + 1


= R.H.S
Question 4.

If A + B + C = π, prove that
Answer:

= sinA + sinB + sinC


Using,




since A + B + C = π



And,





Using , sin2A = 2sinAcosA





And,






= R.H.S
Question 5.

If A + B + C = π, prove that
Answer:

= cosA + cosB + cosC


Using ,




since A + B + C = π



And,





Using , cos2A = 1 – 2sin2A







= R.H.S
Question 6.

If A + B + C = π, prove that
Answer:

= sin2A + sin2B + sin2C


Using,



Sin2A = 2sinAcosA


= 2sinAcosA + 2sin(B+C)cos(B - C)
since A + B + C = π



= 2sinAcosA + 2sin(π - A)cos(B - C )
= 2sinAcosA + 2sinAcos(B - C)
= 2sinA{cosA + cos (B-C)}
( but cos A = cos { 180 - ( B + C ) } = - cos ( B + C )


And now using
= 2sinA{2sinBsinC}
= 4sinAsinBsinC



Now,


= sinA + sinB + sinC


Using,











Therefore,




= R.H.S
Question 7.

If A + B + C = π, prove that

sin (B + C – A) + sin (C + A – B) – sin (A + B – C) = 4cos A cos B sin C
Answer:

= sin (B + C – A) + sin (C + A – B) – sin (A + B – C)


Using,



= 2sinC cos(B-A) – sin(A+B-C)


since A + B + C = π



= 2sinCcos(B-A) – sin(π – C – C)


= 2sinCcos(B-A) – sin2C


Since , sin2A = 2sinAcosA,


= 2sinCcos(B-A) – 2sinCcosC


= 2sinC{cos(B-A) – cosC}


Using ,





= 4cosAcosBsinC


= R.H.S
Question 8.

If A + B + C = π, prove that
Answer:


Taking L.C.M



Multiplying and divide the above equation by 2, we get



Since , sin2A = 2sinAcosA



NOW,


= sin2A + sin2B + sin2C


= 2sinAcosA + 2sin(B+C)cos(B - C)
since A + B + C = π



= 2sinAcosA + 2sin(π - A)cos(B - C )
= 2sinAcosA + 2sinAcos(B - C)
= 2sinA{cosA + cos (B-C)}
( but cos A = cos { 180 - ( B + C ) } = - cos ( B + C )


And now using
= 2sinA{2sinBsinC}
= 4sinAsinBsinC


Putting the above value in the equation, we get



= 2


= R.H.S
Question 9.

If A + B + C = π, prove that

cos2 A + cos2 B + cos2 C = 1 – 2cos A cos B cos C
Answer:

= cos2 A + cos2 B + cos2 C


Using formula ,






Using ,





Using , since A + B + C = π



And, cos(π – A ) = -cosA




Using cos2A = 2cos2A -1




= 1 + cos2A – cosAcos(B-C)


= 1 + cosA{cosA - cos(B-C)}


Using ,




Since , A + B + C = π




= 1 - 2cosAcosCcosC


= R.H.S
Question 10.

If A + B + C = π, prove that

sin2 A – sin2 B + sin2 C = 2sin A cos B sin C
Answer:

= sin2 A – sin2 B + sin2 C


Using formula ,






Using ,





since A + B + C = π



And sin(π – A) = sinA




Using , cos2A = 1 – 2sin2A











= 2sinAcosBsinC


= R.H.S
Question 11.

If A + B + C = π, prove that
Answer:


Using formula ,






Using ,





Using , since A + B + C = π



And, cos(π – A ) = -cosA




Using , cos2A = 1 – 2sin2A




since A + B + C = π


and Using ,





Using , since A + B + C = π




= R.H.S
Question 12.

If A + B + C = π, prove that

tan 2A + tan 2B + tan 2C = tan 2A tan 2B tan 2C
Answer:

= tan 2A + tan 2B + tan 2C


Since A + B + C = π


A + B = π – C


2A + 2B = 2π – 2C


Tan (2A+2B) = tan (2π – 2C)


Since tan (2π – C) = -tan C


Tan (2A + 2B) = -tan 2C


Now using formula,




Tan 2A + tan 2B = -tan 2C + tan 2C tan 2B tan 2A


Tan 2A + tan 2B + tan 2C = tan 2A tan 2B tan 2C


= R.H.S