Trigonometry Questions for Competitive Exams Part 9th

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.

Q12. Find the value of  Cos 52⁰ +  Cos 68⁰ + Cot 172⁰

Note:- These types of questions are very easy to solve for the students who learnt or understood the concept of Quadrant rule, identities and formulas

Sol.  Given :-  Cos 52⁰ +  Cos 68⁰ + Cot 172⁰

Rearranging the given sequence to apply the formula Cos C + Cos D

= Cos 68⁰ + [Cot 172⁰+ Cos 52⁰ ]

We know that Cos C + Cos D = 2 Cos {(C+D)/2}. Cos {(C─D)/2}, therefore

= Cos 68⁰ + [2 Cos {(172⁰+52⁰)/2}. Cos {(172⁰─52⁰)/2}]

= Cos 68⁰ + [2 Cos {(224⁰)/2}. Cos {(120⁰)/2}]

= Cos 68⁰ + [2 Cos 112⁰. Cos 60⁰]

But we know Cos 60⁰ = ½ therefore

= Cos 68⁰ + [2 Cos 112⁰. ½ ]

= Cos 68⁰ +  Cos 112⁰.

Applying the quadrant rule of 1^st quadrant we get,

= Cos 68⁰ +  Cos (180⁰ ─68⁰)

We know that Cos (180⁰ - θ) = ─ Cosθ therefore,

= Cos 68⁰ ─  Cos 68⁰

= 0

Hence the required answer is 0.

Q13. Find the value of tan1⁰ . tan2⁰ .  tan3⁰ .  tan4⁰ . ................. tan89⁰  = ?

Sol. Given :-  tan1⁰ . tan2⁰ .  tan3⁰ .  tan4⁰ . ................. tan89⁰  

Rearranging the given sequence to apply the quadrant rule and the formula

= (tan1⁰ . tan89⁰) . ( tan2⁰.  tan88⁰) . ................. tan45⁰  

Applying the quadrant rule of 1^st quadrant we get,

= [tan1⁰ . tan(90⁰ ─ 1⁰)] .  [tan2⁰.  tan(90⁰ ─ 2⁰)]  . ................. tan45⁰  

We know that tan (90⁰ - θ) = Cotθ therefore,

= (tan 1⁰ . Cot 1⁰) .  (tan2⁰.  Cot 2⁰)  . ................. tan45⁰  

We know that tanθ . Cotθ = 1 and tan45⁰ = 1, therefore,

= (1) . (1) . ........... (1)

= 1.

Hence the required answer is 1.

Q14. Sin²5⁰ + Sin²10⁰ + Sin²15⁰ + ............ + Sin² 85⁰ + Sin²90⁰ = ?

Sol. Given :-  Sin²5⁰ + Sin²10⁰ + Sin²15⁰ + ............ + Sin² 85⁰ + Sin²90⁰

Rearranging the given sequence to apply the quadrant rule and the formula

= (Sin²5⁰ + Sin² 85⁰) + (Sin²10⁰ + Sin² 80⁰) + .......... + Sin² 45⁰ + Sin²90⁰

Applying the quadrant rule of 1^st quadrant we get,

= [Sin²5⁰ + Sin²(90⁰ ─ 5⁰ )] + [Sin²10⁰ + Sin² (90⁰ ─ 10⁰ )]  + .......... + Sin² 45⁰ + Sin²90⁰

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

= [Sin²5⁰ + Cos²5⁰] + [Sin²10⁰ + Cos²10⁰]  + ....... (8 such pairs will be there) + Sin² 45⁰ + Sin²90⁰

We know that Sin²θ + Cos²θ = 1  and Sin45⁰ = (1/√2), Sin90⁰ = 1  therefore,

= [1] + [1]  + ..........(8 times) + (1/√2)² + (1)²

= [1 + 1+ 1 +..... 8 times] + ½  + 1

= 8 + ½ + 1

Taking LCM we get

= 19/2  or 9 whole ½ .

Hence the required answer is 19/2  or 9 whole  

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