This is a second article written for providing guide to CBSE Class 10 students in preparation for final exam. My endeavour in preparing this article has been to explain the subject in the simplest way for the benefit of even the weak students. 

In Mathematics, trigonometry is one of the most important and prominent topics to learn. Trigonometry is basically the study of triangles. CBSE Class 10 Maths curriculum includes the chapter of Trigonometry because it finds applications which would make any serious student of Math or Engineering to sit up. The term ‘Trigon’ means triangle and ‘metry’ means measurement. Trigonometric identities class 10 consists of trigonometry ratios such as sine, cosine and tangent in its equations. Even, trigonometry identities for class 10 formulas are based on these Trigonometry focuses on the relationships between the sides and angles of triangles. It involves various trigonometric functions, and those functions are used for the determination of unknown angles and sides of a triangle. These identities are used to solve various trigonometry problems.

Trigonometry should be understood clearly at its introductory stage to develop a strong base which will prove beneficial in the long run. A trigonometric identity is a trigonometric equation that is true for every possible value of the input variable on which it is defined.

By considering a right-angled triangle, trigonometry identities for class 10 lists could be figured out. The trigonometric identities or equations are formed using trigonometry ratios for all the angles. Using trigonometry identities, we can express each trigonometric ratios in terms of other trigonometric ratios, and if any of the trigonometry ratios value is known to us, then we can find the values of other trigonometric ratios. We can also solve trigonometric identities class 10 questions, using these identities as well.

Trigonometric Identities for Class 10

Trigonometric identities are the equations that include the trigonometric functions such as sine, cosine, tangent, etc., and are true for all values of angle θ. Here, θ is the reference angle taken for a right-angled triangle. In class 10th, there are basically three trigonometric identities, which we learn in the trigonometry chapter. They are:

Cos2 θ + Sin2 θ = 1

1 + tan2 θ = Sec2 θ

1 + Cot2 θ = Cosec2 θ

Here, we will prove one trigonometric identity and will use it to prove the other two. Take an example of a right-angled triangle ΔABC.

Importance of trigonometry and its applications

 

Basics of trigonometry 

The geometrical applications of trigonometry generally focus on the calculation of sides and angles of a right-angled triangle. The exact measurement of an angle is obtained using trigonometry.

There are six trigonometric ratios which depict the relationships between the sides and angles of a right-angled triangle. The ratios for a right-angled triangle ABC are as follows:

  1. Sine (sin) = (perpendicular (opposite side))/Hypotenuse

sin A= BC/AC

  1. cosine (cos)= (Base (adjacent side))/Hypotenuse

  cos A= AB/AC

  1. tangent (tan) = (Perpendicular (opposite side))/ (Base (Adjacent side))

tan A= BC/AB

  1. cotangent (cot) = (Base (adjacent side))/(Perpendicular(opposite side))

cot A= AB/BC

  1. secant (sec) = Hypotenuse/(Base(Adjacent side))

sec A= AC/AB

  1. cosecant (cosec) = Hypotenuse/(Perpendicular(Opposite side))

cosec A= AC/BC

Few important points to note about trigonometry:

  • Each trigonometric ratio is a real number with no unit.
  • Values of trigonometric ratios are always the same for the same angles.

Sometimes a question may ask you to “prove the identity” or “establish the identity.” This is the same idea as when you are asked to show that 

(x -1)2 = x2 – 2x +1

In this type of question, we must show the algebraic manipulations that demonstrate that the left and right side of the equation are in fact equal. You can think of a “prove the identity” problem as a simplification problem where you know the answer: you know what the end goal of the simplification should be and you just need to show the steps to get there.

There are two primary ways to prove identities:

  • Start with a known identity and algebraically manipulate both sides to get a new identity.
  • Start with one side of an identity you want to prove, then manipulate it using known identities to get the other side.

There are no short cuts to success, therefore if you wish to solve math problems of class 10th quickly and effectively all you have to do is practice continuously. Practice certainly makes everything perfect and there is no true gold standard for success than it.

 

Keep practicing and all the best!!

In this article, we will show examples using both strategies. It’s important to note that there is no “standard algorithm” for proving identities! Every problem is a little different, but as you prove more identities you may begin to notice patterns that you can often use (such as writing tangent in terms of sine and cosine).

Something else to keep in mind is that:

  • when we are asked to prove a trigonometric identity, we do not treat the identity like an equation to solve – it isn’t! Instead, we are trying to prove that the two expressions are equal, so we must take care to work with one side at a time rather than applying an operation simultaneously to both sides of the equation.
  • We can also use identities that we have previously learned, like the Pythagorean Identity, while simplifying or proving identities.
  • We can also build new identities from previously established identities. For example, we can come up with another identity if we divide both sides of the Pythagorean Identity by cosine squared (which is allowed since we’ve already shown the identity is true).
  • To write the trigonometric ratios of complementary angles, we consider the following as pairs: (sin, cos), (cosec, sec), and (tan, cot).
  • While writing the trigonometric ratios of supplementary angles, the trigonometric ratio won’t change. The sign can be decided using the fact that only sin and cosec are positive in the second quadrant where the angle is of the form (180-θ).
  • There are 3 formulas for the cos2x formula. Among them, you can remember just the first one because the other two can be obtained by the Pythagorean identity sin2x + cos2x = 1.
  • The half-angle formula of tan is obtained by applying the identity tan = sin/cos and then using the half-angle formulas of sin and cos.

Once you have gone over all the key trig identities in your math class, the next step will be verifying them. Verifying trig identities means making two sides of a given equation identical to each other in order to prove that it is true. You’ll use trig identities to alter one or both sides of the equation until they’re the same.

Verifying trig identities can require lots of different math techniques, including FOIL, distribution, substitutions, and conjugations. The letters FOIL stands for First, Outer, Inner, Last. First means multiply the terms which occur first in each binomial. Each equation will require different techniques, but there are a few tips to keep in mind when verifying trigonometric identities.

When verifying trig identities, keep the following three tips in mind:

  •  Start with the Harder Side

Despite what you may initially want to do, we recommend starting with the side of the equation that looks messier or more difficult. Complicated-looking equations often give you more possibilities to try out than simpler equations, so start with the trickier side so you have more options. 

  •  Remember that you can change both sides

You don’t need to stick to only changing one side of the equation. If you get stuck on one side, you can switch over to the other side and begin changing it as well. Neither side of the equation needs to be the same as how it was originally; as long as both sides of the equation end up being identical, the identity has been verified.

  •  Turn all the Functions into Sines and Cosines

Most students learning trig identities feel most comfortable with sines and cosines because those are the trig functions, they see the most. Make things easier on yourself by converting all the functions to sines and cosines!

You’ll need to have key trig identities memorized in order to do well in your geometry or trigonometry classes. While there may seem to be a lot of trigonometric identities, many follow a similar pattern, and not all need to be memorized.

In this article we have provided solution to some questions which have been asked in previous year’s examination. This will make the students adopt various tricks and techniques to solve other questions.

Question 1:  Prove that 

(1 + cot A – cosec A) (1 + tan A + sec A) = 2

Note: It will convenient to turn all the Functions into Sines and Cosines.

Solution     We have

LHS = (1 + cot A – cosec A) (1 + tan A + sec A)

= (1 + cosAsinA1sinA) (1 + sinAcosA1cosA)  [Turning all the functions in terms of sine and cosine]

[sinA+cosA-1]sinA * [cosA+sinA+1]cosA

=  (sin A+ cos A)2 -1sinA cosA (sin A+ cos A)2                                             

= [(sin2 A + cos2 A) + 2 sin A cos A – 1] / sin A cos A

= [(1 + 2sin A cos A – 1) / sin A cos A

= (2 sin A cos A) / (sin A cos A)

= 2

= RHS

LHS = RHS (proved)

Question 2:     

Prove that 1cosecθ-cotθ1sinθ = 1sinθ –  1cosecθ+cotθ

[ CBSE2010, ’12, ’16, ’18, ‘19]

Hint: We should simplify both sides of the equation separately to prove that both sides are equal.

Solution:

LHS =   1cosecθ-cotθ1sinθ

= 1cosecθ-cotθ x cosecθ+Cotθcosec+cotθ1sinθ    [Multiplying and dividing by {cosec θ+cot θcosec θ+cot θ}]

= cosecθ+Cotθcosec2 θ-cot2 cosec θ [ 1sinθ = cosec] 

= (cosecθ+cotθ) – cosec θ [∵(cosec2 θ-cot2 ) = 1]

= cot θ

RHS = 1sinθ –  1cosec θ+cot θ [ 1sinθ = cosecθ

= cosec θ – 1cosecθ+cotθ xcosec θ-cot θcosec θ-cot θ       [[Multiplying and dividing by {cosec θ-cot θcosec θ-cot θ}]

= cosec θ – cosecθ-cotθcosec2 θ-cot2 [∵(cosec2 θ-cot2 ) = 1

=cosec θ – cosec θ + cot θ

= cot θ

= LHS

∴ LHS = RHS

 

Question 3:  

If cosec θ – sin θ = m, and sec θ – cos θ = n, prove that (m2n)3/2 +((mn2)3/2 = 1

[CBSE 2015]

Solution:

We have m2n = (cosec θ – sin θ)2 * (sec θ – cos θ)

= {(1/sin θ) -sin θ}2 * {(1/cos θ) – cos θ}

= {(1- sin2 θ)/ sin θ} * {(1- cos2 θ)/ cos θ}   [ 1-sin2 θ = cos2 and 1 – cos2 θ = sin θ]

= {(cos2 θ)2 / sin2 θ} * {(sin2 θ) / cos θ}

= (cos4 θ / sin2 θ) * (sin2 θ / cos θ)

= cos3 θ

∴ (m2 n)1/3 = cos θ ……………………………………. (i)

Again, mn2 = (cosec θ – sin θ) * (sec θ – cos θ)2 

      = {(1/sin θ) – sin θ} * {(1/ cos θ) – cos θ}2 

        = {(1 – sin2 θ)/ sin θ} * {(1 – cos2 θ)2 / cos2 θ}

        = (cos2 θ/ sin θ) * (sin4 θ / cos2 θ)

        = (cos2 θ/ cos2 θ) * (sin4 θ/ sin θ)

        = sin3 θ

  ∴ (m2 n)1/3 = sin θ ……………………………… (ii)

On squaring (i) and (ii) and adding, we get

(m2 n)2/3 + (m2 n)2/3 

= cos2 θ + sin2 θ

= 1

(m2n)3/2 +((mn2)3/2 = 1

Proved.

 

Question 4:

If tan x = n tan y and sin x = m sin y, prove that cos2 x = (m2 – 1)/ (n2 – 1)

                                                                                                                                          [CBSE 2019C]

Solution: 

tan x = n tan y

⇒ tan x = n/cot y

⇒ cot y = n/tan x

Again, sin x =m sin y

⇒ sin x = m/ cosec y

⇒ cosec y = m/sin x

Now, cosec2 y – cot2 y = 1

⇒ (m2 / sin2 x) – (n2 / tan2 x) = 1 

⇒ (m2 / sin2 x) – (n2 cos2 x/sin2 x) = 1

⇒ (m2 – n2 cos2 x) = sin2 x

⇒ (m2 – n2 cos2 x) = (1 – cos2 x)

⇒ m2 = 1 – cos2 x + n2 cos2 x

⇒ m2 – 1 = cos2 x (n2 – 1)

⇒ (m2 – 1)/ (n2 – 1) = cos2 x

Hence, cos2 x = (m2 – 1)/ (n2 – 1)

Proved.

 

Some Practice Questions:

Question 1: If a cos θ -b sin θ = c, prove that,

(a sin θ + b cos θ) = ± √ (a2 + b2 + c2)

                                                                                                                                        [CBSE 2015]

 

Question 2: Prove that (sin θ + cosec θ )2 + (cos θ + sec θ)2 = 7 + tan2 θ + cot2 θ.

Question 3: Prove that (1 + cot A – cosec A) (1 + tan A + sec A) = 2.
                                                                                                                                        [CBSE 2019]

Question 4: Prove that: (sin A – 2 sin3 A)/ (2 cos3 A – cos A) = tan A                   [CBSE 2018]

Question 5: Prove that: sec2 θ – (sin2 θ – 2 sin4 θ)/ (2 cos4 θ – cos2 θ) = 1       [CBSE 2012]