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.
Q15. Find the value of tan40 . tan430 . tan470
. tan860 = ?
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 :-
tan40 . tan430 . tan470 . tan860
Rearranging
the given sequence to apply the Quadrant rule
= tan40
. tan860 . tan430 . tan470
Applying
the quadrant rule of 1st quadrant we get,
= tan40
. tan(900 ─ 40) . tan430 . tan(900 ─
430)
We know
that tan (900 - θ) = Cotθ therefore,
= (tan40
. Cot 40) . (tan430 . Cot 430)
We know
that tanθ . Cotθ = 1, therefore,
= (1) . (1)
= 1.
Hence the
required answer is 1.
Q16. Sin(500 + θ) ─
Cos(40 ─ θ) = ?
Sol. Given
:- Sin(500 + θ) ─ Cos(40 ─ θ)
Applying
the quadrant rule of 1st quadrant we get,
= Sin[900
─ (400 ─ θ)] ─ Cos(40 ─ θ)
We know
that Sin (900 - θ) = Cosθ
therefore,
= Cos (400
─ θ) ─ Cos(40 ─ θ)
= 0.
Hence the
required answer is 0.
Q17. Find the value of tan4050.
Sol. Given
:- tan4050.
Applying
the quadrant rule we get,
= tan(3600
+ 450) {this
will lie in the 1st Quadrant}
We know
that tan (3600 + θ) = tanθ
therefore,
= tan450
= 1.
Hence the
required answer is 1.
Q18. Find the value of Sin2250.
Sol. Given
:- Sin2250.
Applying
the quadrant rule we get,
= Sin(1800
+ 450) {this
will lie in the 3rd Quadrant}
We know
that Sin (1800 + θ) = ─ Sinθ
therefore,
= ─ Sin450
= (─1/√2).
Hence the
required answer is (─1/√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........................
Thank you .
ReplyDelete