To Know More About Maths Trigonometry Formulas And Problems

Introduction of trigonometry

Trigonometry is the branch of mathematics which deals with relations involving lengths and angles of a triangle. Simply it can be known as the study of triangles. The angles are measured in degrees or radians.

Triangle and their types

A triangle consisting of three-sides three-angles and up to 180 degrees. And also a triangle is a well defined bounded figure consisting of all these. On the basis of length, it is divided into 3 types: Scalene(all the sides and angles are totally different), isosceles(2sides and 2angles are equal) and equilateral triangle(all the sides and angles are equal). On the basis of angles, it is again divided into three kinds:acute-angled triangle(all the angles are less than 90degress), right-angled triangle(one angle is equal to 90 degrees) and obtuse-angled triangle(one angle is greater than 90 degrees).

Some basic points of trigonometry chapter

  • A triangle is equal when the sum of the angles of a triangle is equal to 180 degrees.
  • The sum of the length of the two sides of a triangle is greater than the third side whereas the difference between the lengths of two sides is less than the third side of the length.
  • The longest side of a triangle is equal to the largest angle of a triangle.
  • Area of a triangle =½ baseheight.

So now we are going to discuss a few formulas and problems related to the trigonometry chapter.

  • Sin2A+cos2A=1
  • Sin2A=1-cos2A
  • Sin A=√1-cos2A
  • Cos2A=1-sin2A
  • Cos A=√1-sin2A
  • Sec2A-tan2A=1
  • Sec2A=1+tan2A
  • SecA=√1+tan2A
  • Tan2A=sec2A-1
  • TanA=√sec2A-1
  • Cosec2-cot2A=1
  • Cosec2A=1+cot2A
  • CosecA=√1+cot2A
  • Cot2A=cosec2A-1
  • CotA=√cosec2A-1
  • Sin A=1-cos2A
  • Sin(A+B)=sinA. cosB+cosA. sinB
  • Cos(A+B)=cosA.cosB-sinA.sinB
  • Sin(A-B)=sinA.cosB-cosA.sinB
  • Cos(A-B)=cosA.cosB+sinA.sinB
  • Sin(A+B).sin(A-B)=sin2A-sin2B=cos2B-cos2A
  • Cos(A+B).cos(A-B)=cos2A-sin2B=cos2B-sin2A
  • Tan(A+B)= TanA+tanB/1-tanA.tanB
  • Tan(A-B)=TanA-tanB/1+tanA.tanB
  • Cot(A+B)=CotA.cotB-1/cotA+cotB
  • Cot(A-B)=CotA.cotB+1/cotB-cotA
  • Sin(90-)=cos
  • Cos(90-)=sin
  • Tan(90-)=cot
  • Cot(90-)=tan
  • Sec(90-)=cos
  • Cosec(90-)=sin
  • Sin(90+)=+cos
  • Cos(90+)=-sin
  • Tan(90+)=-cot
  • Cot(90+)=-tan
  • Sec(90+)=+sec
  • Cosec(90+)=-cosec
A 30° 45° 60° 90°
SinA 0 1/2 1/2 3/2 1
CosA 1 √3/2 1/√2 1/2 0
TanA 0 1/√3 1 √3 Not defined
CosecA Not defined 2 √2 2/√3 1
SecA 1 2/√3 √2 2 Not defined
CotA Not defined √3 1 1//√3 0

These all are the formulas related to the trigonometry chapter so now let us discuss a few problems related to this chapter.

1. Sin2θ=cos2θ, then find tan5 θ?
Sol: Sin2 θ=cos2 θ, tan5 θ
Cos(90-2 θ) =cos3 θ
90=3 θ+2 θ
90=5 θ
Tan5 θ=tan90
2. If cosec θ+cot θ=k then prove that cos θ=k2-1/k2+1?
Sol: Cos θ=k2-1/k2+1
Given that,
Cosec θ+cot θ=k
We know that
Cosec2 θ-cot2 θ=1

a2-b2=(a+b) (a-b)

(cosec θ+cot θ) (cosec θ-cot θ)=1
K. (cosec θ-cot θ)=1
cosec θ-cot= θ1/k
Cosec θ+cot θ=k
Cosec θ-cot θ=k
2cosec θ=k=1/k
2cosec θ=k2+1/k

3. Sin15; cos15
Sol: Sin15=sin(45-30)
Sin(A-B)=sinA cosB- cosA sinB
= Sin45cos30-cos45sin30
= 1/√2 X √3/2 – 1/√2X1/2
= √3

4. Simplify secA(1-sinA) (SecA+tanA)?
Sol: SecA(1-sinA) (SecA+tanA)
= (secA-sinA-secA) (secA+tanA)
=(secA-sinA 1/cosA) (secA+tanA)
=(secA-sinA/cosA) (secA+tanA)
= (secA-tanA) (secA+tanA)
= sec2A-tan2A
= 1
5. If sec θ+tan θ=p, then what is the value of sec θ-tan θ?
Sol: Given that,
sec θ+tan θ=p
According trigonometric identity
sec2 θ -tan2 θ =1
(sec θ – tan θ) (sec θ – tan θ)=1
p=(sec θ – tan θ)=1

sec θ – tan θ = 1

6. Find sin2A cos θ 2A?
Sol: sinA-sinA+cosA sinA
= 2sinA cosA
Therefore, sin2A
= cos2A-cos(A+A)
= cosA cosA-sinA sinA
Therefore, cos2A-cos2A-sin2A

Cos2 A. 1-2sin2A


These are some of the trigonometry formulas and problems which are very important and useful in higher studies. These are very important and it helps in many of the competitive exams. By learning them we can gain a lot of knowledge.

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