## Trigonometric Ratios

In this page trigonometric ratios we are going to see all of the formulas based on the sides.

Let us consider the following right triangle.

In the triangle below the side which is opposite to 90 degree is called the hypotenuse side (AC). The side which is opposite to theta is called the opposite side(AB) and the remaining side is called the adjacent side (BC).

In maths we have six trigonometric ratios.They are sin θ,cos θ,tan θ,cosec θ,sec θ and cot θ • sin θ = Opposite side/Hypotenuse side
• cos θ = Adjacent side/Hypotenuse side
• tan θ = Opposite side/Adjacent side
• Cosec θ = Hypotenuse side/Opposite side
• Sec θ = Hypotenuse side/Adjacent side
• cot θ = Adjacent side /Opposite side

Now let us see the reciprocal formula

• sin θ = 1/Cosec θ
• Cosec θ = 1/sin θ
• Cos θ = 1/sec θ
• sec θ = 1/cos θ
• tan θ = 1/cot θ
• cot θ = 1/tan θ

Example problem based on the above formula

Example 1
Show that (1+tan θ) /(1+cot θ) = (sin θ + tan θ)/(1 + cosθ)

L.H.S

=     (1 + tan θ) /(1 + cot θ)

=     (1+tan θ) /[1+(1/tan θ)]

=     (1+tan θ) /[(tan θ + 1)/tan θ)]

=      (1+tan θ) x[tan θ)/(tan θ + 1)]

=      tan θ --------(1)

R.H.S

=      (sin θ + tan θ)/(1 + cos θ)

=      [sin θ + (sin θ/cos θ)]/(1+cos θ)

=      [(sin θ cos θ + sin θ)/cos θ ]/ (1+cos θ)

=      [sin θ(cosθ+1)/ cosθ] / (1+cos θ)

=      [sin θ (1+cos θ) / cos θ] x 1/[(1+cosθ)]

=      sin θ/cosθ

=      tan θ --------(2)

Both (1) and (2) are equal

Hence it is proved.

Example 2

Prove that (tan A + cot B)/(cot A + tan B) = tan A/tan B

L.H.S

=   (tan A + cot B)/(cot A + tan B)

=   [tan A + (1/tan B)]/[(1/tan A) + tan B)]

=   [(tan A tan B + 1)/tan B]/[(1+tan A tan B) /tan B]

=   [(tan A tan B + 1)/tan B] x [tan B/(1+tan A tan B)]

=   tan A/tan B

R.H.S

Hence it is proved.

You can find more practice questions to get clear idea about this topic.

Related Topics

Trigonometric Ratios to trigonometry 