Trigonometry Questions for Competitive Exams Part 14th

Following the sequence of last post on basics of Trigonometry I am continuing the basics with their related questions in this post. Let’s prepare this to get full hold on the questions of trigonometry.

Some More Very Important Trigonometric Formulae useful for competitive Exams:-

Sin3θ = 3Sinθ – 4Sin3θ Cos3θ = 4Cos3θ – 3Cosθ Tan3θ = (3tanθ – 3tan3θ)/(1 + 3tan2θ) Sinθ . Sin2θ . Sin4θ = ¼ Sin3θ. Cosθ . Cos2θ . Cos4θ = ¼ Cos3θ. Tanθ . Tan2θ  . Tan4θ = Tan3θ. Maximun value of m.Sinθ ± n.Cosθ = √(m2 + n2) Maximun value of m.Sinθ ± n.Sinθ = √(m2 + n2) Maximun value of m.Cosθ ± n.Cosθ = √(m2 + n2)

Q28. If tan(A + B) = ½  and tan(A - B) = 1/3 then find the value tan2A?

Sol.  Given :-  tan(A + B) = ½ 

tan(A - B) = 1/3

Let A + B = X  ........................................ eq.no.1

&   A -  B = Y  ......................................... eq.no.2

From equations 1 & 2 (On adding we get)

2A = (X + Y)

Tan2A = tan (X + Y)

Using tan2θ formula we get,

Tan 2A = (tanX + tanY)/(1 – tanX. tanY)

Putting the values of X and Y we get,

Tan2A = [tan(A + B) + tan(A - B)]/[1 – tan(A + B) . tan(A -B)]

Putting the values of tan(A + B) & tan(A - B) we get,

Tan2A = [(1/2) + (1/3)]/ [1 – (1/2).(1/3)]

Tan2A = {5/6}/{1 – (1/6)}

Tan2A = (5/6)/(5/6)

Tan2A = 1.

Hence the required answer is Tan2A = 1.

Q29. If A + B = 45⁰ then, find the value of (1 + tanA). (1 + tanB)

Sol. Given A + B = 45⁰

Tan(A + B) = tan45⁰

{tanA + tanB}/ {1 – tanA . tanB} = 1

tanA + tanB = 1 – tanA . tanB

tanA + tanB + tanA . tanB = 1

Adding 1, in both sides we get,

1 + tanA + tanB + tanA . tanB = 1 + 1

1 (1 + tanA) + tanB (1 + tanA) = 2.

(1 + tanA). (1 + tanB) = 2.

Hence the required answer is (1 + tanA). (1 + tanB) = 2.

Q30. Find the value of Sin(π/18) . Sin(5π/18) . Sin(7π/18) .

Sol. Given :-  Sin(π/18) . Sin(5π/18) . Sin(7π/18) .

We know that π = 180⁰ therefore,

= Sin(180/18) . Sin{(5 × 180)/18} . Sin{(7 × 180)/18}.

= Sin10⁰ . Sin50⁰ . Sin70⁰ .

= Sin(90⁰ - 80⁰) . Sin(90⁰ - 40⁰) . Sin(90⁰ - 20⁰)

We know that Sin(90⁰ - θ) = Cosθ therefore,

= Cos80⁰. Cos40⁰ . Cos20⁰.

= Cos20⁰ . Cos40⁰ . Cos80⁰.

We know that, Cosθ . Cos2θ . Cos4θ = ¼ Cos3θ. Therefore,

= ¼ . Cos(3 × 20⁰)

= ¼ . Cos60⁰.

= ¼ . ½

= 1/8 .

Hence the required answer Sin(π/18) . Sin(5π/18) . Sin(7π/18) = 1/8 .

 (Note you may say that this is so long but dear this is the only way to explain the whole concept of the question to you now its upto you that how fast and short handedly you can solve these type of questions)     

                                                                  

Hope you have enjoyed the base concept of this chapter till now. More Conceptual enjoyment will be continued for you soon, just keep visiting this blog regularly and writing me your related queries through comment or forum.

Best of Luck for Exams. To be continued soon........................