**Evaluate the Given Trigonometric Function by Trigonometric Ratios :**

In this section, you will learn how to evaluate trigonometric functions using trigonometric ratios.

**Question 1 :**

If cos A = 3/5, then find the value of

(sin A - cos A)/2 tan A

**Solution :**

**cos A = 3/5 = Adjacent side / Hypotenuse side**

(Opposite side)^{2} = (Hypotenuse side)^{2} - (Adjacent side)^{2}

= 5^{2} - 3^{2}

(Opposite side)^{2} = 25 - 9 = 16

Opposite side = √16 = 4

sin A = opposite side/ hypotenuse side = 4/5

tan A = opposite side / Adjacent side

tan A = 4/3

(sin A - cos A)/2 tan A = (4/5) - (3/5) / 2(4/3)

= (1/5) / (8/3)

= 3/40

**Question 2 :**

If cos A = 2x/(1 + x^{2}), then find the values of sin A and tan A in terms of x.

**Solution :**

**cos A = 2x/(1 + x ^{2}) = Adjacent side / Hypotenuse side**

**Opposite side ^{2} = **

** = 1 + x ^{4} + 2x^{2} - 4x^{2}**

** = 1 + x ^{4} - 2x^{2}**

** = (1 - x ^{2})^{2}**

**Opposite side = 1 - x ^{2}**

**sin A = Opposite side / Hypotenuse side**

**sin A = **** (1 - x ^{2})/**

**tan A = Opposite side / Adjacent side**

**tan A = **** (1 - x ^{2})/**

**Question 2 :**

If sin θ = a/√a^{2} + b^{2}, then show that

b sin θ = a cos θ

**Solution :**

**Given that, sin θ = a/√a ^{2} + b^{2 }**

**sin ****θ**** = Opposite side / Hypotenuse side**

**(Adjacent side) ^{2} = (Hypotenuse)^{2} - (Opposite side)^{2}**

**(Adjacent side) ^{2} = (**

**Adjacent side = b**

**b sin θ = ****ab/√a ^{2} + b^{2 } ----(1)**

**cos ****θ = b/****√a ^{2} + b^{2}**

**a cos**** ****θ = ab/****√a ^{2} + b^{2 }----(2)**

**(1) = (2)**

**Hence proved.**

**Question 3 :**

If 3cot A = 2 , then find the value of

(4sin A - 3cos A)/(2sin A + 3cos A)

**Solution :**

cot A = 2/3 = Adjacent side / Opposite side

(Hypotenuse side)^{2 } = (Opposite side)^{2} + (Adjacent side)^{2}

(Hypotenuse side)^{2 } = 3^{2} + 2^{2}

Hypotenuse side = **√13**

**sin A = Opposite side / Hypotenuse side**

sin A = 3/**√13**

**cos A = Adjacent side / Hypotenuse side**

**cos A = 2**/**√13**

**(4 sin A - 3 cos A)/(2 sin A + 3 cos A) :**

** = [4(3/****√13****) - 3(2/****√13)****]****/[2(3/****√13****) ****+ 3(2/****√13****)]**

** = [(12-6)****/****√13****]****/[(6+6)/****√13****]**

** = 6****/12**

** = 1/2**

After having gone through the stuff given above, we hope that the students would have understood how to evaluate the trigonometric Function using trigonometric ratios.

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