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If (cos θ + sin θ) = √2 cos θ, Show that (cos θ – sin θ) = √2 sin θ

  • Last Updated : 21 Sep, 2021

Trigonometry is the relation between the angles and sides of a right-angled triangle. In a right angles triangle, there are 3 angles of which one angle is a right angle (90°) and the other two angles are acute angles and there are 3 sides. The side opposite to the right angle is called Hypotenuse. There are 6 ratios between these sides based on the angle between them and they are called Trigonometric Ratios.

The 6 Trigonometric ratios are:

  • Sine (sin)
  • Cosine (cos)
  • Tangent (tan)
  • Cosecant (cosec)
  • Secant (sec)
  • Cotangent (cot)

Right-angled triangle ACB

Sine (sin):

Sine of an angle is defined by the ratio of lengths of sides which is opposite to the angle and the hypotenuse. For the above triangle, the value of angle sine is given for both ∠A and ∠B, the definition for sine angle is the ratio of the perpendicular to its hypotenuse.

SinA =  \frac{P}{H}= \frac{CB}{AB}

SinB = \frac{P}{H}= \frac{AC}{AB}

Cosine (cos):

Cosine of an angle is defined by the ratio of lengths of sides which is adjacent to the angle and the hypotenuse. For the above triangle, the value of the angle cos is given for both ∠A and ∠B, the definition for cos angle is the ratio of the base to its hypotenuse.

CosA = \frac{B}{H}= \frac{AC}{AB}

CosB = \frac{B}{H}= \frac{CB}{AB}

Tangent (tan):

The tangent of an angle is defined by the ratio of the length of sides which is opposite to the angle and the side which is adjacent to the angle. For the above triangle, the value of the angle tan is given for both ∠A and ∠B, the definition for tan angle is the ratio of the perpendicular to its base.

TanA = \frac{P}{B}= \frac{CB}{AC}

TanB = \frac{P}{B}= \frac{AC}{CB}

Cosecant (cosec):

The cosecant of an angle is defined by the ratio of the length of the hypotenuse and the side opposite the angle. For the above triangle, the value of the angle cosec is given for both ∠A and ∠B, the definition for cosec angle is the ratio of the hypotenuse to its perpendicular.

CosecA =  \frac{H}{P}= \frac{AB}{CB}

CosecB =  \frac{H}{P}= \frac{AB}{AC}

Secant (sec):

Secant of an angle is defined by the ratio of the length of the hypotenuse and the side and the side adjacent to the angle. For the above triangle, the value of the angle sec is given for both ∠A and ∠B, the definition for sec angle is the ratio of the hypotenuse to its base.

SecA =  \frac{H}{P}= \frac{AB}{AC}

SecB =  \frac{H}{P}= \frac{AB}{CB}

Cotangent (cot):

The cotangent of an angle is defined by the ratio of the length of sides that is adjacent to the angle and the side which is opposite to the angle. For the above triangle, the value of the angle cot is given for both ∠A and ∠B, the definition for cot angle is the ratio of the hypotenuse to its base.

CotA = \frac{B}{P}= \frac{AC}{CB}

CotB = \frac{B}{P}= \frac{CB}{AC}

If (cos θ + sin θ) = √2 cos θ, Show that (cos θ – sin θ) = √2 sin θ

Solution:

cosθ + sinθ = √2 cosθ                           ……… ( 1 )

Squaring both sides, we get

cos2θ + 2.cosθ.sinθ + sin2θ = 2.cos2θ

\therefore  2.cosθ.sinθ = 2.cos2θ – sin2θ – cos2θ

\therefore  2.cosθ.sinθ = cos2θ – sin2θ

\therefore  2.cosθ.sinθ = (cosθ – sinθ)(cosθ + sinθ)

From (1), we get

2.cosθ.sinθ = (cosθ – sinθ)(√2 cosθ)

\therefore  (cosθ – sinθ)(√2) = 2.sinθ

\therefore  (cosθ – sinθ) = \frac{\large2.sinθ}{\large√2}

\therefore  (cosθ – sinθ) = √2 sinθ

Hence, Proved.

Similar Problems

Question 1: Verify that: 1/(sin θ + cos θ) + 1/(sin θ – cos θ) = 2 sin θ/(1 – 2 cos2 θ)

Solution:

L.H.S\frac{\large1}{\large(sin θ + cos θ)}  \frac{\large1}{\large(sin θ - cos θ)}

\therefore  L.H.S = \frac{\large(sinθ - cosθ) + (sinθ + cosθ)}{\large(sinθ + cosθ)(sinθ - cosθ)}

\therefore  L.H.S = \frac{\large2sinθ}{\large(sin^2θ - cos^2θ)}

Since, sin2θ + cos2θ = 1

\therefore  L.H.S = \frac{\large2sinθ}{\large(1 - cos^2θ - cos^2θ)}

\therefore  L.H.S = \frac{\large2sinθ}{\large(1 - 2cos^2θ)}

\therefore  L.H.S = R.H.S

\therefore  \frac{\large1}{\large(sin θ + cos θ)}  \frac{\large1}{\large(sin θ - cos θ)}  = \frac{\large2sinθ}{\large(1 - 2cos^2θ)}

Hence, Proved.

Question 2: Prove the identity: sinθ / (1 + cosθ) = (1 − cosθ) / sinθ

Solution:

L.H.S = \frac{\large sinθ}{\large(1 + cosθ)}

Multiplying numerator and denominator by (1 – cosθ);

L.H.S = \frac{\large sinθ(1 - cosθ)}{\large(1 + cosθ)(1 - cosθ)}

\therefore  L.H.S = \frac{\large sinθ - sinθ.cosθ}{\large(1 - cos^2θ)}

Since, sin2θ + cos2θ = 1 ⇒  sin2θ = 1 – cos2θ 

\therefore  L.H.S = \frac{\large sinθ(1 - cosθ)}{\large sin^2θ}

\therefore  L.H.S = \frac{\large (1 - cosθ)}{\large sinθ}

\therefore  L.H.S = R.H.S

\therefore  \frac{\large sinθ}{\large(1 + cosθ)}  = \frac{\large (1 - cosθ)}{\large sinθ}

Hence, Proved.

Question 3: Prove the trigonometric equation: cos θ.sec θ/cot θ = tan θ.

Solution:

L.H.S = \frac{\large cos θ.sec θ}{\large cot θ}

Since, \large secθ = \frac{\large 1}{\large cosθ}

\therefore  L.H.S = \frac{\large cos θ.\frac{\large 1}{\large cos θ}}{\large cot θ}

\therefore  L.H.S = \frac{\large 1}{\large cot θ}

Since, \large tanθ = \frac{\large 1}{\large cotθ}

\therefore  L.H.S = tanθ

\therefore  L.H.S = R.H.S

\therefore \frac{\large cos θ.sec θ}{\large cot θ} =\large tanθ

Hence, Proved.


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