Question

Solve pls kindly don not spam​

Answer 1

Step-by-step explanation:

★ Concept :-

Here the concept of Trigonometry has been used. We see that we have to prove that L.H.S. is equal to R.H.S. here. We see that in R.H.S. there is a single term so we need to simplify L.H.S. only here. For simplifying L.H.S. firstly we have yo draw a triangle and then apply ratio of sides there. Then uwe can simplify ir and thus apply Pythagoras Theorem there to prove the answer.

Let's do it !!

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★ Solution :-

Given to prove,

$</p><p>\;\;\bf{\mapsto\;\;\green{\sin^{2}x\;+\;\cos^{2}x\;=\;1}}$

From here we get,

$\;\;\sf{\leadsto\;\;\blue{L.H.S.\;=\;\sin^{2}x\;+\;\cos^{2}x}}$

$\;\;\sf{\leadsto\;\;\orange{R.H.S.\;=\;1}}$

Let's understand the Triangle given in attachment of the answer

There we have took the reference angle as x.

We know that,

>> sin x = Perpendicular / Hypotenuse

>> cos x = Base / Hypotenuse

Now let's apply these ratios in the L.H.S.

Then we get,

$\;\;\tt{\rightarrow\;\;L.H.S.\;=\;\bigg(\dfrac{Perpendicular}{Hypotenuse}\bigg)^{2}\;+\;\bigg(\dfrac{Base}{Hypothetenuse}\bigg)^{2}}$

$\;\;\tt{\rightarrow\;\;L.H.S.\;=\;\dfrac{(Perpendicular)^{2}}{(Hypotenuse)^{2}}\;+\;\dfrac{(Base)^{2}}{(Hypothetenuse)^{2}}}$

Since the denominators are same. So we can easily add them.

$\;\;\tt{\rightarrow\;\;L.H.S.\;=\;\dfrac{(Perpendicular)^{2}\;+\;(Base)^{2}}{(Hypotenuse)^{2}}}$

By Pythagoras Theorem, we know that in a right angled triangle,

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

(Hypotenuse)² = (Base)² + (Perpendicular)²Here we are dealing with a right angled triangle since trignometry is applicable in right angled triangle only.

So by applying this, we get

$\;\;\tt{\rightarrow\;\;L.H.S.\;=\;\dfrac{(Hypotenuse)^{2}}{(Hypotenuse)^{2}}}$

$\;\;\bf{\rightarrow\;\;\red{L.H.S.\;=\;1}}$

Clearly L.H.S. = R.H.S.

So,

$\;\;\bf{\rightarrow\;\;\purple{R.H.S.\;=\;L.H.S.\;=\;1}}$

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★ More to know :-

• sec² A = 1 + tan² A

• cosec² A = 1 + cot² A

• cosec A = 1/(sin A)

• sec A = 1/(cos A)

• cot A = 1/(tan A)

Answer 2

Answer:

Hope it will help you

Step-by-step explanation:

Then:

sin

θ

=

a

c

cos

θ

=

b

c

So:

sin

2

θ

+

cos

2

θ

=

a

2

c

2

+

b

2

c

2

=

a

2

+

b

2

c

2

By Pythagoras

a

2

+

b

2

=

c

2

, so

a

2

+

b

2

c

2

=

1

So given Pythagoras, that proves the identity for

θ

∈

(

0

,

π

2

)

For angles outside that range we can use:

sin

(

θ

+

π

)

=

−

sin

(

θ

)

cos

(

θ

+

π

)

=

−

cos

(

θ

)

sin

(

−

θ

)

=

−

sin

(

θ

)

cos

(

−

θ

)

=

cos

(

θ

)

So for example:

sin

2

(

θ

+

π

)

+

cos

2

(

θ

+

π

)

=

(

−

sin

θ

)

2

+

(

−

cos

θ

)

2

=

sin

2

θ

+

cos

2

θ

=

1

Pythagoras theorem

Given a right angled triangle with sides

a

,

b

and

c

consider the following diagram:

enter image source here

The area of the large square is

(

a

+

b

)

2

The area of the small, tilted square is

c

2

The area of each triangle is

1

2

a

b

So we have:

(

a

+

b

)

2

=

c

2

+

4

⋅

1

2

a

b

That is:

a

2

+

2

a

b

+

b

2

=

c

2

+

2

a

b

Subtract

2

a

b

from both sides to get:

a

2

+

b

2

=

c

2