Question

ask any 10 trignometry proving​

Answer 1

Answer:

PLZ THANK

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.

Trigonometric Identities Class 10 Proof

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;

(1). AB2 + BC2 = AC2

Dividing equation (1) by AC2, we get,

AB2AC2 + BC2AC2 = AC2AC2 (ABAC)2 + (BCAC)2 = (ACAC)2 (Cosθ)2+(Sinθ)2 = 1

(2). Cos2 θ + Sin2 θ = 1

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 = AC2AB2 (ABAB)2 + (BCAB)2 = (ACAB)2

(3). 1 + Tan2 θ = Sec2 θ

If θ = 0, then,

1 + tan20 = sec20

1 + 02 = 12

1 = 1

And if we put θ = 90,then

1 + tan290 = sec290

1 + ∞ = ∞

∞ = ∞

As you can see both the sides values are equal. Therefore, it proves that for all the values between 00 and 900, the equation (3) is satisfied. So, it is also a trigonometric identity.

Let’s see what we get if we divide equation (1) by BC2, we get,

AB2BC2 + BC2BC2 = AC2BC2 (ABBC)2 + (BCBC)2 = (ACBC)2

(4). Cot2 θ + 1 = Cosec2 θ

Now let us prove this identity as well.

If θ = 0, then equation (4) can be written as;

Cot20 + 1 = Cosec20

∞ + 1 = ∞

∞ = ∞

Both the sides are equal.

And if θ = 90,then equation (4) can be written as;

Cot290 + 1 = Cosec290

02 + 1 = 12

1 = 1

Thus proved that equation (4) is a trigonometric identity.

PLZ THANK