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Class 10 Maths Concepts for chapter Trignometry





What is Trigonometry

Trigomometry from Greek trigõnon, "triangle" and metron, "measure") is a branch of mathematics that studies relationships involving lengths and angles of triangles. The field emerged during the 3rd century BC from applications of geometry to astronomical studies.
Trigonometry is most simply associated with planar right angle triangles (each of which is a two-dimensional triangle with one angle equal to 90 degrees). The applicability to non-right-angle triangles exists, but, since any non-right-angle triangle (on a flat plane) can be bisected to create two right-angle triangles, most problems can be reduced to calculations on right-angle triangles. Thus the majority of applications relate to right-angle triangles

Right triangle trigonometry and Trigonometric ratios


In a right angle triangle ABC where B=90° ,
trigonometric ratios
We can define following term for angle A
trigonometric ratios
We can define the trigonometric ratios for angle A as
trigonometric ratios
Notice that each ratio in the right-hand column is the inverse, or the reciprocal, of the ratio in the left-hand column.
The reciprocal of sin A is cosec A ; and vice-versa.
The reciprocal of cos A is sec A
And the reciprocal of tan A is cot A
These are valid for acute angles.
We can define tan A = sin A/cos A
And Cot A =cos A/ Sin A
Important Note
Since the hypotenuse is the longest side in a right triangle, the value of
sin A or cos A is always less than 1 (or, in particular, equal to 1).
Similarly we can have define these for angle $C$
trigonometric ratios
We can define the trigonometric ratios for angle $C$  as
sin C= Perpendicular/Hypotenuse =AB/AC
cosec C= Hypotenuse/Perpendicular =AC/AB
cos C= Base/Hypotenuse =BC/AC
sec C= Hypotenuse/Base=AC/BC
tan C= Perpendicular/Base =AB/BC
cot C= Base/Perpendicular=BC/AB

Trigonometric Ratios of Common angles


We can find the values of trigonometric ratio’s various angle
trig reference table
Question 1
State whether the following are true or false. Justify your answer.
(i) The value of tan A is always less than 1.
(ii) sec A =12/5 for some value of angle A.
(iii) cos A is the abbreviation used for the cosecant of angle A.
(iv) tan A is the product of tan and A.
(v) sin ? =5/3 for some angle
Solution
i) False . The value of tan A increase from 0 to infinity
ii) True . The value of sec A increase from 1 to infinity
iii) False .Cos A is the abbreviation used for the cosine of angle A
iv) False .cot A is one symbol. We cannot separate it
v) False. The value of sin ? always lies between 0 and 1 and 5/3 > 1
Question 1
The value of (sin30 + cos30) (sin60 + cos60) is (A) 1 (B) 0 (C) 1 (D) 2 Solution
Answer (B)

Trigonometric ratio’s of complimentary angles


We know that for Angle A, the complementary angle is 90 – A
In a right angle triangle ABC
A+B+C=180
Now B=90
So A +C =90
Or C=90-A
We have see in Previous section the value for  trigonometric ratios for angle $C$
sin C= Perpendicular/Hypotenuse =AB/AC
cosec C= Hypotenuse/Perpendicular =AC/ABB
cos C= Base/Hypotenuse =BC/AC
sec C= Hypotenuse/Base=AC/BC
tan C= Perpendicular/Base =AB/BC
cot C= Base/Perpendicular=BC/AB
This can be rewritten as
sin (90-A)  =AB/AC
cosec (90-A)  =AC/AB
cos (90-A) =BC/AC
sec (90-A)=AC/BC
tan (90-A)=AB/BC
cot( 90-A) =BC/AB
Also we know that
sin A=BC/AC
cosec A= AC/BC
cos A= AB/AC
sec A =AC/AB
tan A =BC/AB
cot A =AB/BC
From both of these, we can easily make it out
Sin (90-A) =cos(A)
Cos(90-A) = sin A
Tan(90-A) =cot A
Sec(90-A)= cosec A
Cosec (90-A) =sec A
Cot(90- A) =tan A

Trigonometric identities


We have studied pythagorus theorem in earliar classes which states that for a right-angled triangle the square on the hypotenuse is equal to the sum of the squares on the other two side
If a is the hypthonues and b and c are other two sides,then
a2= b2 + c2

This same theorem can be used to proved the below trigonometric identities
Sin2 A + cos2 A  =1
1 + tan2 A =sec2 A
1 + cot2 A =cosec2 A

How to solve Trigonometric identities Problems


1) Learn well the formulas for Trigonometric identities, trigonometric ratios,reciprocals The better you know the basic identities, the easier it will be to recognise what is going on in the problems. 2) Generally RHS( Right hand side) would be more complex. So start from there and simplify it to the same form as LHS(Left hand side) 3) It becomes many times easy, if Convert all sec, csc, cot, and tan to sin and cos. Most of this can be done using the quotient and reciprocal identities. 4) If you see power 2 or more, it will involve using the below identities mostly Sin2 A + cos2 A  =1
1 + tan2 A =sec2 A
1 + cot2 A =cosec2 A
5) Once you get the hang of it, you will begin to see patterns and it will be easy to solve these Trigonometric identities Problems 6) Practice and Practice. You will soon start figuring out the equation and there symmetry to resolve them fast
Question 1
Prove that (sin4 θ – cos4θ +1) cosec2θ = 2
Solution
L.H.S. = (sin4 θ – cos4θ +1) cosec2θ
= [(sin2θ – cos2θ) (sin2θ + cos2θ) + 1] cosec2θ
= (sin2θ – cos2θ + 1) cosec2θ
[Because sin2θ + cos2θ=1]
= 2sin2θ cosec2θ [Because 1– cos2θ = sin2θ ]
= 2 = RHS

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