# Proving Trigonometric Ratios

In proving trigonometric ratios we will learn how to proof the questions step-by-step using trigonometric identities.

1. If sin4x + sin2 x = 1 then proof that, cot4x + cot2 x = 1.

Solution:

Given, sin4x + sin2 x = 1

⇒ sin4x + sin2 x - sin2 x = 1 - sin2 x, [subtract sin2 x from both the sides]

⇒ sin4x = 1 - sin2 x

⇒ sin4x = cos2 x.

Now L.H.S. = cot4x + cot2 x

= cos4x/sin4x + cos2 x/sin2 x

= cos4x/cos2 x + sin4x/sin2 x, [since, sin4x = cos2 x and cos2 x = sin4x]

= 1 = R.H.S. [since we know, sin2 x + cos2 x = 1]

(Proved)

2. If sin θ - cos θ = √2 cos θ then proof that sin θ + cos θ = √2 sin θ, where 0 < θ < π/2

Solution:

Given, sin θ - cos θ = √2 cos θ

⇒ (sin θ - cos θ)2 = (√2 cos θ)2, [squaring both the sides]

⇒ sin2 θ + cos2 θ - 2 sin θ cos θ = 2 cos2 θ

⇒ sin2 θ + cos2 θ - 2 sin θ cos θ - cos2 θ = 2 cos2 θ - cos2 θ, [subtract cos2 θ from both the sides]

⇒ sin2 θ - 2 sin θ cos θ = cos2 θ

⇒ sin2 θ - 2 sin θ cos θ + 2 sin θ cos θ = cos2 θ + 2 sin θ cos θ, [adding 2 sin θ cos θ on both the sides]

⇒ sin2 θ = cos2 θ + 2 sin θ cos θ

⇒ sin2 θ + sin2 ϴ = sin2 θ + cos2 θ + 2 sin θ cos θ, [adding sin2 θ on both the sides]

⇒ 2 sin2 θ = (sin θ + cos θ)2

⇒ (sin θ + cos θ)2 = 2 sin2 θ

Now taking square root on both the sides we get,

⇒ sin θ + cos θ = ± √2 sin θ

According to the question, 0 < θ < π/2, hence we neglect the negative vaue.

Therefore, sin θ + cos θ = √2 sin θ

(Proved)

The above explanation on proving trigonometric ratios will help us to solve different types of trigonometric problems.

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Trigonometric Functions