Maharashtra State Board 11th Maths Solutions Chapter 3 Trigonometry – II Ex 3.5
Question 1
Maharashtra Board Solution
In Δ ABC, A + B + C = π, show that cos2A + cos2B + cos2C = – 1 – 4 cosA cosB cosC
Solution & Step-by-Step Answer:
L.H.S. = cos 2A + cos 2B + cos 2C = = 2.cos(A + B).cos (A – B) + 2cos2C – 1 In ΔABC, A + B + C = π ∴ A + B = π – C ∴ cos(A + B) = cos(π – C) ∴ cos(A + B) = – cosC ………….(i)
∴ L.H.S. = – 2.cos C.cos (A – B) + 2.cos2C – 1 …[From(i)]
= – 1 – 2.cosC.[cos(A – B) – cosC]
= – 1 – 2.cos C.[cos(A – B) + cos(A + B)]
… [From (i)]
= – 1 – 2.cos C.(2.cos A.cos B)
= – 1 – 4.cos A.cos B.cos C = R.H.S.
Question 2
Maharashtra Board Solution
sin A + sin B + sin C = 4 cos A/2 cos B/2 cos C/2
Solution & Step-by-Step Answer:
Question 3
Maharashtra Board Solution
cos A + Cos B + Cos C = 4 cos A/2 cos B/2 cos C/2 =
Solution & Step-by-Step Answer:
L.H.S. = sin A + sin B + sin C = In Δ ABC, A + B + C = π, ∴ A + B = π – C
Question 4
Maharashtra Board Solution
sin2 A + sin2 B – sin2 C = 2 sin A sin B cos C
Solution & Step-by-Step Answer:
We know that, sin2 = L.H.S. = sin2 + sin2 B + sin2 C = 1 – cos(A + B). cos(A – B) – sin2C = (1 – sin2 C ) – cos (A + B). cos (A – B) = cos2 C – cos(A + B). cos(A – B) ∴ cos(A + B) = cos(it — C) ∴ cos(A + B) = — cos C …(i) ∴ L.H.S. = cos2C + cos C.cos(A – B) … [From (i)] = cos C[cos C + cos(A – B)] = cos C[- cos(A + B) + cos(A – B)] … [From (i)] = cos C[cos (A-B) – cos(A + B)] = cos C(2 sin A.sin B) = 2 sin A.sin B. cos C = R.H.S. [Note: The question has been modified.]
Question 5
Maharashtra Board Solution
=
Solution & Step-by-Step Answer:
Question 6
Maharashtra Board Solution
tan tan tan tan tan tan = 1
Solution & Step-by-Step Answer:
In Δ ABC, A + B + C = π ∴ A + B = π – C
Question 7
Maharashtra Board Solution
Solution & Step-by-Step Answer:
In Δ ABC, A + B + C = π ∴ A + B = π – C
Question 8
Maharashtra Board Solution
tan 2A + tan 2B + tan 2C = tan 2A tan 2B + tan 2C
Solution & Step-by-Step Answer:
In Δ ABC, A + B + C = π ∴ 2A + 2B + 2C = 2π ∴ 2A + 2B = 2π – 2C tan(2A + 2B) = tan(2n — 2C) = -tan 2C ∴ tan2A+tan2B=—tan2C.(1-tan2A.tan2B) ∴ tan 2A + tan 2B = – tan2C+ tan2A.tan2B.tan2C ∴ tan 2A + tan 2B + tan 2C = tan2A.tan2B.tan2C
Question 9
Maharashtra Board Solution
cos2 A + cos2 B – cos2 C = 1 – 2 sin A sin B sin C
Solution & Step-by-Step Answer:
we know that cos2θ = L.H.S. = cos2 A + cos2 B + cos2 C = 1 + cos (A + B).cos(A — B) – cos2 C In ΔABC, A + B + C = π A + B = π — C cos(A + B) = cos(π — C) cos(A + B) = -cosC ………….. (i) L.H.S. = 1 — cos C.cos(A — B) — cos2 C …[From(i)] = 1 — cos C.[cos(A — B) + cos C] = 1 — cos C.[cos(A — B) — cos(A + B)]...[From (i)] = 1 — cos C.(2.sin A.sin B) = 1 — 2.sinA.sin B.cos C = R.H.S.
