Trigonometry Questions for Competitive Exams Part 6th

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.

Q3. Find the value of Cos²30⁰ . Cos²45⁰ + 4Sec²60⁰ + ½ Cos²90⁰ – 2tan²60⁰.

Sol. These types of questions are very easy to solve for the students, who had learnt or know the concept of preparation of trigonometric ratio table for various angles

Given :-  Cos²30⁰ . Cos²45⁰ + 4Sec²60⁰ + ½ Cos²90⁰ – 2tan²60⁰.

We now through trigonometric table Cos30⁰ = √3/2

                                                                  Cos45⁰ = 1/√2

                                                                  Sec60⁰ = 2

                                                                  Cos90⁰ = 0

                                                                  Tan60⁰ = √3

Putting these values in the given data to evaluate its value:-

= (√3/2)² × (1/√2)² + 4 × (2)² + ½ (0) – 2 (√3)²

= (¾ × ½) + (4 × 4) – (2 × 3)

= (3/8) + 10

= 83/8

Hence the required value of the given question is 83/8.

Q4. If Sinθ – Cosθ = 0 then find the value of Sin⁴θ + Cos⁴θ.

Sol. Given :-  Sinθ – Cosθ = 0

                        Sinθ = Cosθ

Dividing both sides by Cosθ we get,

                        Tanθ = 1

                        Tanθ = tan45⁰

                             θ = 45⁰

Now,   Sin⁴θ + Cos⁴θ

Putting the value of θ in the above data we get

= (1/√2)⁴ + (1/√2)⁴

= ¼  + ¼  

= ½ .

Hence the required anwer of the question is ½

Q5. If Sinθ + Cosecθ =2 then find the value of Sin²θ + Cosec²θ.

Note :- These type of question are a type of question which disguise the students. Actually these questions are based on algebraic formula (x + 1/x)² = x² + 1/x² + 2

We know that  Cosecθ = 1/Sinθ hence the equation can be written as

Sinθ + (1/Sinθ) = 2 and if Sinθ = x then this equation will take the form (x + 1/x) =2

Point to Remembered: - If (x + 1/x) = 2

                                           Then xⁿ + (1/xⁿ) = 2

                                          Where n = 1, 2, 3, 4, .............. any natural number

Sol.  Method 1^st :-

Given Equation :- Sinθ + Cosecθ =2

Squaring both sides:- (Sinθ + Cosecθ)² = (2)²

                                        Sin²θ + Cosec²θ + 2× Sinθ × Cosecθ = 4

But we know that Sinθ × Cosecθ =1

              Therefore        Sin²θ + Cosec²θ + 2× 1 = 4

                                        Sin²θ + Cosec²θ = 4 – 2 =2

Hence required answer  Sin²θ + Cosec²θ =2

Method 2^nd :- As per rule:-

                                            If (x + 1/x) = 2

                                           Then xⁿ + (1/xⁿ) = 2

                                          Where n = 1, 2, 3, 4, .............. any natural number

We can answer this type of question direct verbally within second that

 Sin²θ + Cosec²θ =2      (because Cosecθ = 1/Sinθ )

Hence required answer  Sin²θ + Cosec²θ =2

 (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........................