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Prove that 1/sec A – tan A – 1/cosA = 1/cos A – 1/sec A + tan A

  • Last Updated : 29 Oct, 2021

Trigonometry is a discipline of mathematics that studies the relationships between the lengths of the sides and angles of a right-angled triangle. Trigonometric functions, also known as goniometric functions, angle functions, or circular functions, are functions that establish the relationship between an angle to the ratio of two of the sides of a right-angled triangle. The six main trigonometric functions are sine, cosine, tangent, cotangent, secant, or cosecant.

Angles defined by the ratios of trigonometric functions are known as trigonometry angles. Trigonometric angles represent trigonometric functions. The value of the angle can be anywhere between 0-360°.

Right angled triangle

As given in the above figure in a right-angled triangle:

  • Hypotenuse: The side opposite to the right angle is the hypotenuse, It is the longest side in a right-angled triangle and opposite to the 90° angle.
  • Base: The side on which angle C lies is known as the base.
  • Perpendicular: It is the side opposite to angle C in consideration.

Trigonometric Functions

Trigonometry has 6 basic trigonometric functions, they are sine, cosine, tangent, cosecant, secant, and cotangent. Now let’s look into the trigonometric functions. The six trigonometric functions are as follows,

  • sine: It is defined as the ratio of perpendicular and hypotenuse and It is represented as sin θ
  • cosine: It is defined as the ratio of base and hypotenuse and it is represented as cos θ
  • tangent: It is defined as the ratio of sine and cosine of an angle. Thus the definition of tangent comes out to be the ratio of perpendicular and base and is represented as tan θ
  • cosecant: It is the reciprocal of sin θ and is represented as cosec θ.
  • secant: It is the reciprocal of cos θ and is represented as sec θ.
  • cotangent: It is the reciprocal of tan θ and is represented as cot θ.

According to the above image, Trigonometric Ratios are

Sin θ = Perpendicular / Hypotenuse = AB/AC

Cosine θ = Base / Hypotenuse = BC / AC

Tangent θ = Perpendicular / Base = AB / BC

Cosecant θ = Hypotenuse / Perpendicular = AC/AB

Secant θ = Hypotenuse / Base = AC/BC

Cotangent θ = Base / Perpendicular = BC/AB

Reciprocal Identities

Sin θ = 1/ Cosec θ                    OR        Cosec θ = 1/ Sin θ



Cos θ = 1/ Sec θ                       OR        Sec θ = 1 / Cos θ

Cot θ = 1 / Tan θ                     OR         Tan θ = 1 / Cot θ

Cot θ = Cos θ / Sin θ               OR         Tan θ = Sin θ / Cos θ

Tan θ.Cot θ = 1

Some other identities are 

sin2x + cos2x = 1

1 + tan2x = sec2x

1 + cot2x = cosec2x

Trigonometric Identities of Complementary and Supplementary Angles

  • Complementary Angles: Pair of angles whose sum is equal to 90°
  • Supplementary Angles: Pair of angles whose sum is equal to 180°

Identities of Complementary angles are



sin (90° – θ) = cos θ

cos (90° – θ) = sin θ

tan (90° – θ) = cot θ

cot (90° – θ) = tan θ

sec (90° – θ) = cosec θ

cosec (90° – θ) = sec θ

Identities of supplementary angles

sin (180° – θ) = sin θ

cos (180° – θ) = – cos θ

tan (180° – θ) = – tan θ

cot  (180° – θ) = – cot θ

sec (180° – θ) = – sec θ

cosec (180° – θ) = – cosec θ

Values of Trigonometric Ratios

 30°45°60°90°
sin θ0    1/2               1/√2            √3/2            1
cos θ1√3/21/√210
tan θ01/√31√3Not Defined
sec θNot Defined2√22/√31
cosec θ12/√3√22Not Defined
cot θNot Defined√311/√30

Quadrants of trigonometry

Prove that 1/sec A – tan A – 1/cosA = 1/cos A – 1/sec A + tan A

Solution: 

We have 1/sec A – tan A – 1/cosA = 1/ cos A -1/sec A + tan A

First take LHS 

1/ (sec A – tan A) – 1/cosA 



multiply and divide 1/ (sec A – tan A) by (sec A + tan A)

so now we can writre as 

= {1/ (sec A – tan A) × (sec A + tan A)/(sec A + tan A)} – 1/cosA 

= {(sec A + tan A)/ (sec2 A – tan2 A)} – 1/cosA 

= {(sec A + tan A)/ 1} – sec A                        {1 + tan2x = sec2x or sec2x – tan2x = 1}

= sec A + tan A – sec A                                {Sec θ = 1 / Cos θ }

= tan A

Now RHS

1/ cos A -1/sec A + tan A

multiply and divide {1/ (sec A + tan A)} by (sec A – tan A)

Now we can write as 

= 1/ cos A – [{1/ (sec A + tan A)} × (sec A – tan A )/(sec A – tan A)]  

= 1/ cos A – [(sec A – tan A) / (sec2 A – tan2 A)]                 {1 + tan2x = sec2x   or  sec2x – tan2x = 1}

= sec A – [(sec A – tan A) / 1]                                            {Sec θ = 1 / Cos θ}

= sec A    –  sec A + tan A

= tan A

Therfroe LHS = RHS 

So, 1/sec A – tan A – 1/cos A = 1/ cos A – 1/sec A + tan A

Hence proved

Similar Questions

Question 1: Prove cot2θ – 1/sin2θ = -1 



Solution:

We have  cot2θ – 1/sin2θ

= cot2θ –  cosec2θ                                     { Cosec2θ = 1/ Sin2 θ }

= – (cosec2θ – cot2θ)                                                  

= -1 

= RHS 

Hence Proved

Question 2: Prove (1 + cot2θ) (1 – cos θ)(1 + cos θ) = 1

Solution:

We have LHS 

= (1 + cot2θ) (1 – cos θ)(1 + cos θ) 

= (1 + cot2θ) (1 – cos2 θ)

= cosec2θ sin2θ                                   {1 + cot2θ = cosec2θ  and 1 – cos2 θ  = sin2θ}

= (1/sin2θ) × sin2θ                              {Cosec2θ = 1/ Sin2θ}

= 1

= RHS

Hence Proved


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