Trigonometry Basics for Competitive Exams Part 4th

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.

Concept 1^st :- Basic Questions based on Trigonometric Ratios:-

Q1. If tanθ = 8/15 and θ is acute angle then find cosecθ.

Sol. In this type of question we draw a figure of right angled triangle and use the Geometry theorem “Pythagoras Theorem” to solve this type of question.

Right angled triangle

In the figure triangle ABC, right angled at B and acute angle θ at C

                       AB = Perpendicular

                       BC = Base

                       AC = Hypotenuse

Given Data:- tanθ = (8/15)

We know that tanθ = (Perpendicular/Base) = (AB/BC) = (8/15)

Let AB = 8k and BC = 15k.

According to Pythagoras theorem (AC)² = (AB)² + (BC)²

Putting the values we get :-              (AC)² = (8k)² + (15k)²

On solving we get :-                           (AC)² = 289k²

Taking square root both sides :-        AC = 17k

Now, Cosecθ = (Hypotenuse/perpendicular) = (AC/AB) = 17k/8k

Therefore required answer Cosecθ = (17/8).

Q2. Evaluate :- (3Cos²30⁰ + Sec²30⁰ + 2Cos0⁰ +  3Sin90⁰ – tan²60⁰)

Sol. In this type of question we have the values of trigonometric ratios from the table which I have given in my last post

Putting the values from table:-  3(√3/2)² + (2/√3)² + 2(1) + 3(1) - (√3)²

                                                        = 3(3/4) + (4/3) + 2 + 3 - (3)

                                                        = (9/4) + (4/3) + 2

On taking LCM and adding we get:- = (67/12)

According to law of fraction:-

 divide this fraction and convert it in mix fraction = 5whole(7/12)

Hence the required answer is (67/12) or [5whole(7/12)]

Q3. If 3tanθ = 4 then find the value of (3Sinθ + 2Cosθ)/(3Sinθ - 2Cosθ)

Sol. This type of Question is solved in the following pattern:-

Dividing numerator and denominator by Cosθ, we get

             [(3Sinθ + 2Cosθ)/Cosθ]/[(3Sinθ - 2Cosθ)/Cosθ]

         = [(3Sinθ/Cosθ + 2Cosθ/Cosθ)]/[( 3Sinθ/Cosθ - 2Cosθ/Cosθ)]

We know that (Sinθ/Cosθ) = tanθ  therefore

          = [(3tanθ + 2)/ (3tanθ - 2)]

But according to question 3tanθ = 4, therefore putting this value we get

          = [(4 + 2)/(4 - 2)]

          = [6/2] = 3

Hence the required answer will be 3.

(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........................