Question

how to solve trigonometry lesson for class 10th​

Answer 1

Answer:

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Step-by-step explanation:

Trigonometric identities class 10 includes basic identities of trigonometry. When we recall, an equation is considered as an identical, if the equations are true for all the values of variables involved. Similarly, trigonometric equation, which involves trigonometry ratios of all the angles, is called a trigonometric identity if it is true for all values of the angles.

In Mathematics, trigonometry is one of the most important and prominent topics to learn. Trigonometry is basically the study of triangles. The terms ‘Trigon’ means triangle and ‘metry’ means measurement. Trigonometric identities class 10 consists of trigonometry ratios such as sine, cosine and tangent in its equations. Even, trigonometry identities class 10 formula are based on these ratios. These identities are used to solve various trigonometry problems.

By considering a right-angled triangle, trigonometry identities class 10 lists could be figured out. The trigonometric identities or equations are formed using trigonometry ratios for all the angles. Using trigonometry identities, we can express each trigonometric ratios in terms of other trigonometric ratios, and if any of the trigonometry ratios value is known to us, then we can find the values of other trigonometric ratios. We can also solve trigonometric identities class 10 questions, using these identities as well.

Trigonometric Identities for Class 10

In class 10th, there are basically three trigonometric identities, which we learn in trigonometry chapter. They are:

Cos2 θ + Sin2 θ = 1

1 + Tan2 θ = Sec2 θ

1 + Cot2 θ = Cosec2 θ

Here, we will prove on trigonometric identity and will use it to prove the other two. Take an example of a right-angled triangle ΔABC.



Proof of Trigonometric Identities Class 10

In a right-angled triangle, by the Pythagorean theorem, we know,

(Perpendicular)2 + (Base)2 = (Hypotenuse)2

Therefore, in ΔABC, we have;

AB2 + BC2 = AC2 ….. (1)

Dividing equation (1) by AC2, we get,

AB2AC2 + BC2AC2 = AC2AC2

 

(ABAC)2 + (BCAC)2 = (ACAC)2

 

(Cosθ)2+(Sinθ)2 = 12

Cos2 θ + Sin2 θ = 1 …..(2)

If θ = 0, then,

Cos2 0 + Sin2 0 = 1

12 + 02 = 1

1 + 0 = 1

1 = 1

And if we put θ = 90,then

Cos2 90 + Sin2 90 = 1

02 + 12 = 1

0 + 1 = 1

1 = 1

For all angles, 0°≤ θ ≤ 90°, equation (2) is satisfied. Hence, equation (2) is a trigonometric identity.

Again, divide equation (1) by AB2, we get

AB2AB2 + BC2AB2 = AC2A