# If a sin θ – b cos θ = c then prove that a cos θ + b sin θ = ± √(a^{2} + b^{2} – c^{2}).

**Trigonometry **is basically the study of the relationship between the angles and the sides of a triangle. It is one of the widely used topics of mathematics that is used in daily life. It involves operations on a right-angled triangle i.e. a triangle having one of the angles equal to 90°. There are some terms that we should know before going further. These terms are,

**Hypotenuse**– It is the side opposite to the right angle in a right-angled triangle. It is the longest side of a right-angled triangle. In figure1 side, AC is the hypotenuse.**Perpendicular**– The perpendicular of a triangle, corresponding to a particularly acute angle θ is the side opposite to the angle θ. In figure1 side, AB is the perpendicular corresponding to angle θ.**Base**– It is the side adjacent to a particularly acute angle θ. In figure 1 side BC is the base corresponding to angle θ.

As earlier said, trigonometry depicts the relationship between the angles and sides of a right-angled triangle. This relationship is represented by standard ratios and is given as follows:

**Sine (sin):**The sine of an angle θ is the ratio of the length of the perpendicular, corresponding to the angle θ, to the length of the hypotenuse of the triangle.

sin θ = perpendicular/hypotenuse = p/h

**Cosine (cos):**The cosine of an angle θ is the ratio of the length of the base, corresponding to the angle θ, to the length of the hypotenuse of the triangle.

cos θ = base/hypotenuse = b/h

**Tangent (tan):**The tangent of an angle θ is the ratio of the length of the perpendicular, corresponding to the angle θ, to the length of the base for the particular angle of the triangle.

tan θ = perpendicular/base = p/b

**Cotangent (cot):**it is the reciprocal of a tangent.

cot θ = 1/tan θ = base/perpendicular = b/p

**Secant (sec):**it is the reciprocal of cosine.

sec θ = 1/cos θ = hypotenuse/base = h/b

**Cosecant (cosec):**it is the reciprocal of sine.

cosec θ = 1/sin θ = hypotenuse/perpendicular = h/p

**Trigonometric functions of Complementary angles**

One of the relations of trigonometry includes the concept of complementary angles. Complementary angles are a set of two angles, say x and y, such that on adding them they evaluate to 90°. Therefore, we can say x = 90° – y. There exist a special complementary relation between the trigonometric ratios as given below

**Between sin and cos: **

sin(90° – x) = cos x

cos(90° – x) = sin x

**Between tan and cot:**

tan(90° – x) = cot x

cot(90° – x) = tan x

**Between sec and cosec:**

sec(90° – x) = cosec x

cosec(90° – x) = sec x

**Trigonometric Indentities**

These identities depend on the Pythagorean theorem. Applying Pythagoras theorem to the right-angled triangle, we get:

Opposite^{2} + Adjacent^{2} = Hypotenuse^{2}

Dividing both sides by Hypotenuse^{2}

Opposite^{2}/Hypotenuse^{2} + Adjacent^{2}/Hypotenuse^{2} = Hypotenuse^{2}/Hypotenuse^{2}

sin

^{2}x + cos^{2}x = 11 + tan

^{2}x = sec^{2}x1 + cot

^{2}x = cosec^{2}x

In the question given below, some complementary relationships between the trigonometric ratios will be used.

**If a sin** θ** – b cos** θ** = c , prove that a cos **θ** + b sin** θ** = ± √(a**^{2} + b^{2} – c^{2})

^{2}+ b

^{2}– c

^{2})

**Solution:**

a sin θ – b cos θ = c

Squaring both sides:

⇒(a sin θ – b cos θ)

^{2}= c^{2}⇒a

^{2}sin^{2}θ + b^{2}cos^{2}θ – 2ab sinθ cosθ = c^{2}Using the property that sin

^{2}θ = 1-cos^{2}θ and cos^{2}θ = 1-sin^{2}θ⇒a

^{2}(1-cos^{2}θ) + b^{2}(1-sin^{2}θ) – 2ab sinθ cosθ = c^{2}⇒a

^{2 }– a^{2}cos^{2}θ + b^{2}– b^{2}sin^{2}θ – 2ab sinθ cosθ = c^{2}Moving a

^{2}and b^{2}to right hand side⇒-a

^{2}cos^{2}θ – b^{2}sin^{2}θ – 2ab sinθ cosθ = c^{2}-a^{2}-b^{2}Multiplying by -1 on both sides

⇒a

^{2}cos^{2}θ + b^{2}sin^{2}θ + 2ab sinθ cosθ = a^{2}+b^{2}-c^{2}Now if we observe left hand side is square of (a cosθ + b sin θ)

⇒(a cosθ + b sin θ)

^{2 }= a^{2}+b^{2}-c^{2}Taking Square roots on both sides

a cosθ + b sin θ = ± √(a

^{2}+ b^{2}– c^{2})Hence proved!

**Similar Problems**

**Question 1: If θ be an acute angle and 7sin2θ + 3cos2θ = 4, then find the value of tanθ.**

**Solution: **

7sin2θ + 3cos2θ = 4

7sin2θ + 3(1-sin2θ) = 4

7sin2θ + 3 – 3sin2θ = 4

4sin2θ = 1

sin2θ = 1/4

sinθ = 1/2

So, θ = 30o

then tanθ = 1/√3

**Question 2: If cosα = a cosβ and sinα = b sinβ, then find the value of sin2β in terms of a and b.**

**Solution: **

On squaring both sides

cos

^{2}α = a^{2}cos^{2}β=> 1 – sin

^{2}α = a^{2}(1 – sin^{2}β) ……..(1)Again, sinα = b sinβ

On squaring both sides

sin

^{2}α = b^{2}sin^{2}βPut the value of sin

^{2}α in (1)1 – b

^{2}sin^{2}β = a^{2}– a^{2}sin^{2}β)a

^{2}– 1 = a^{2}sin^{2}β – b^{2}sin^{2}βa

^{2}– 1 = sin^{2}β(a^{2}– b^{2})sin

^{2}β = a^{2}– 1 / (a^{2}– b^{2})

**Question 3: a, b, c are the lengths of three sides of a triangle ABC. If a, b, c are related by the relation a ^{2} + b^{2} + c^{2} **=

**ab + bc + ca, then find the value of (sin**

^{2}A + sin^{2}B + sin^{2}C).**Solution: **

Acc. to question

a

^{2}+ b^{2}+ c^{2}– ab – bc – ca = 0=>2a

^{2}+ 2b^{2}+ 2c^{2}– 2ab – 2bc – 2ca = 0=>(a-b)

^{2}+ (b-c)^{2}+ (c-a)^{2}= 0=> a=b=c

All three sides are equal then it is equilateral triangle.

then ∠A = ∠B = ∠C = 60°

So, sin

^{2}60 + sin^{2}60 + sin^{2}60= 3(√3/2)

^{2}= 9/4

**Question 4: If cot θ = 7/8, evaluate**

**(i) ((1 + sinθ) * (1 – sinθ))/(1 + cosθ) * (1 – cosθ)))**

**(ii) cot2θ**

**Solution:**

(i) Using (a + b) * (a – b) = a

^{2}– b^{2}in numerator and denominatorWe get

(1 – sin

^{2}θ)/(1 – cos^{2}θ)Using sin

^{2}θ + cos^{2}θ = 1We get

cos

^{2}θ/sin^{2}θ = cot^{2}θNow

cot

^{2}θ = (7/8)^{2}= 49/64(ii) cot

^{2}θ = (7/8)^{2}= 49/64

**Question 5: If 3 cot A = 4, Check whether (1 – tan ^{2}A)/(1 + tan^{2}A) = cos^{2}A – sin^{2}A**

**Solution:**

We know that, tanA = sinA / cosA ….(1)

Using (1) on L.H.S

= (1 – sin

^{2}A/cos^{2}A)/(1 + sin^{2}A/cos^{2}A)which on rearranging becomes

= (cos

^{2}A – sin^{2}A)/(cos^{2}A + sin^{2}A)Using the identity,

cos

^{2}A + sin^{2}A = 1LHS becomes

= (cos

^{2}A – sin^{2}A)This is equal to RHS.

LHS = RHS (for every value of cot A)

Hence, Proved.