# If tan A = (sin θ – cos θ)/(sin θ + cos θ) then prove that sin θ + cos θ = ± √2 cos A

Trigonometry is a branch of mathematics that deals with trigonometric ratios that are used to determine the angles and unknown sides of a triangle. Angles are scaled in either radians or degrees. Angles in trigonometry that are widely exercised are 0°, 30°, 45°, 60°, and 90°.

**Trigonometric ratios**

Trigonometric ratios are 6 basic trigonometric relations that form the basics of trigonometry. These 6 trigonometric relations are ratios of all the distinguishable achievable combinations in a right-angled triangle and the angles are represented using the mathematical symbol θ (theta).

The six basic trigonometric ratios are:

- Sine
- Cosine
- Tangent
- Cosecant
- Secant
- Cotangent

Reciprocals of Trigonometric Ratios:

- sin θ = 1/cosec θ
- cos θ = 1/sec θ
- tan θ = 1/cot θ

**Trigonometric Identity**

Trigonometric Identity is an equation that contains trigonometric ratios of an angle if it’s true for all values of the angle. These are usable whenever trigonometric functions are involved in an expression or an equation. All these trigonometric ratios are outlined using the sides of the right triangle, similar to an adjoining side, opposite side, and hypotenuse side.

Some of the identities are:

- tan θ = sin θ/cos θ
- cot θ = cos θ/sin θ
- tan θ . cot θ = 1
- sin
^{2}θ + cos^{2}θ = 1 - 1 + tan
^{2}θ = sec^{2}θ - 1 + cot
^{2}θ = cosec^{2}θ

### To Prove: sin θ + cos θ = ± √2 cos A, where given that tan A = (sin θ – cos θ)/(sin θ + cos θ)

**Solution:**

We have, tan A = (sin θ – cos θ)/(sin θ + cos θ)……(i)

As we know, 1 + tan

^{2 }A = sec^{2}A……(ii)On substituting the value of tan A from (i) to (ii)

=> 1 + {(sin θ – cos θ)/(sin θ + cos θ)}

^{2 }= sec^{2}A=> 1 + (sin θ – cos θ)

^{2}/(sin θ + cos θ)^{2}= sec^{2}A=> {(sin θ + cos θ)

^{2}+ (sin θ – cos θ)^{2}}/(sin θ + cos θ)^{2 }= sec^{2}A=> (sin

^{2}θ + cos^{2}θ + 2 sin θ cos θ + sin^{2}θ + cos^{2}θ – 2 sin θ cos θ)/(sin θ + cos θ)^{2}= sec^{2}A=> (1 + 1)/(sin θ + cos θ)

^{2}= sec^{2}A [ ∵ sin^{2}θ + cos^{2}θ = 1 ]=> 2/(sin θ + cos θ)

^{2}= sec^{2}A=> 2/sec

^{2}A = (sin θ + cos θ)^{2}=>

^{ }2 cos^{2 }A = (sin θ + cos θ)^{2}[ ∵ cos^{2}A = 1/sec^{2}A ]=> (± √2 cos A)

^{2}= (sin θ + cos θ)^{2}=> ± √2 cos A = sin θ + cos θ [ performing square root in both sides ]

∴ sin θ + cos θ = ± √2 cos A,

Hence Proved.

### Similar Questions

**Question 1: If sin θ + cos θ = √2 cos θ (θ ≠ 90°), then find the value of tan θ**

**Solution: **

Given sin θ + cos θ = √2 cos θ

=> (sin θ + cos θ)/cos θ = √2 cos θ/cos θ [ dividing both sides by cos θ ]

=> (sin θ/cos θ) + (cos θ/cos θ) = √2

=> tan θ + 1 = √2

=> tan θ = √2 – 1

**Question 2: If sin θ + cos θ = √2** **cos θ, then show that cos θ − sin θ = √2 sin θ**

**Solution: **

Given: sin θ + cos θ = √2 cos θ

To Prove: cos θ − sin θ = √2 sin θ

Proof:-

sin θ + cos θ = √2 cos θ

=> sin

^{2}θ + cos^{2}θ + 2 sin θ cos θ = 2 cos^{2}θ [ squaring both sides ]=> sin

^{2}θ – cos^{2}θ + 2 sin θ cos θ = 0=> – sin

^{2}θ – cos^{2}θ + 2 sin θ cos θ = – 2 sin^{2}θ [ substracting both sides by 2 sin^{2}θ ]=> sin

^{2}θ + cos^{2}θ – 2 sin θ cos θ = 2 sin^{2}θ=> (cos θ – sin θ)

^{2}= 2 sin^{2}θ=> cos θ − sin θ = √2 sin θ, Hence Proved.

**Question 3:** **If xcos θ + y sin θ = p and x sin θ – y cos θ = q, then find x ^{2} + y^{2}**

**Solution:**

p

^{2}= x^{2}cos^{2}θ + y^{2}sin^{2}θ + 2 xy sin θ cos θq

^{2}= x^{2}sin^{2}θ + y^{2}cos^{2}θ – 2 xy sin θ cos θp

^{2}+ q^{2}= x^{2}cos^{2}θ + y^{2}sin^{2}θ + 2 xy sin θ cos θ + x^{2}sin^{2}θ + y^{2}cos^{2}θ – 2 xy sin θ cos θp

^{2}+ q^{2}= x^{2}(sin^{2}θ + cos^{2}θ) + y^{2}(sin^{2}θ + cos^{2}θ) = x^{2}+ y^{2}[ ∵ sin^{2}θ + cos^{2}θ = 1 ]∴ x

^{2}+ y^{2}= p^{2}+ q^{2}

**Question 4:** **The value of 2sin ^{6} θ + 2cos^{6} θ + 6 sin^{2} θ cos^{2} θ is **

**Solution:**

We have, 2sin

^{6}θ + 2cos^{6}θ + 6 sin^{2}θ cos^{2}θ= 2( sin

^{6}θ + cos^{6}θ + 3 sin^{2}θ cos^{2}θ (sin^{2}θ + cos^{2}θ))= 2((sin

^{2}θ)^{3}+ (cos^{2}θ)^{3}+ 3 sin^{2}θ cos^{2}θ (sin^{2}θ + cos^{2}θ))= 2((sin

^{2}θ + cos^{2}θ)^{3})= 2 x 1

^{3}= 2

**Question 5: If sin θ + sin ^{2} θ = 1, then the value of cos^{12} θ + 3 cos^{8} θ + cos^{6} θ is**

**Solution:**

Given, sin θ + sin

^{2}θ = 1=> sin θ = 1 – sin

^{2}θ=> sin θ = cos

^{2}θ……(i)Let, (sin

^{2}θ + cos^{2}θ)^{3}= 1^{3}=> sin

^{6}θ + cos^{6}θ + 3 sin^{2}θ cos^{2}θ (sin^{2}θ + cos^{2}θ) = 1=> (cos

^{2}θ)^{6}+ cos^{6}θ + 3 (cos^{2}θ)^{2}cos^{2}θ = 1=> cos

^{12}θ + cos^{6}θ + 3 cos^{8}θ = 1∴ The value of cos

^{12}θ + 3 cos^{8}θ + cos^{6}θ is 1.