NCERT Solutions Class 10 Maths Chapter 8 Exercise 8.4 is going to guide you through the most crucial exercise in Chapter 8. The exercise generally addresses the problems involved in trigonometric identities, a stepping stone towards the intricate concepts in later classes. The exercise focuses on the connection between trigonometric identities and various trigonometric ratios. We will continue with this important concept, which will benefit you in your board exams as well as in later math courses.
Trigonometric identities are formulas that include trigonometric ratios (sine, cosine, tangent, etc.) of an angle and hold for all the values of the concerned angles.These are basic mathematical formulas that help simplify expressions and work out problems that require understanding the interrelations among different trigonometric functions. In this exercise, we shall be studying some of the fundamental trigonometric identities and their proofs.
The exercise mentions three main trigonometric identities for the right-angled triangle. These identities will continuously be used in the exercise to solve various problems of trigonometry.
One of the most significant and basic trigonometric identities is the Pythagorean identity used in exercise 8.4.
This identity comes from the Pythagorean theorem when applied to a right-angled triangle. According to the theorem, the sides of a right-angled triangle share the following relation:
Here, AB, BC, and AC represent the sides of the triangle, with AC being the hypotenuse.
After dividing both sides of the above equation by AC2, it can be re-written as the following Pythagorean identity of trigonometry:
In this identity, A is the angle formed opposite to the perpendicular of the triangle such that, 0°A90°. The condition mentioned here holds true for all angles A.
This identity connects the secant and tangent functions. It states the result of the Pythagorean theorem by dividing both sides of the formula by the square of the right triangle's adjacent side. The identity states that the square of the secant of an angle equals one plus the square of the tangent of the same angle. This identity comes in handy when simplifying expressions and solving equations containing these two functions in exercise.
In this identity, 0°A＜90°.
The cosecant-cotangent identity relates the cosecant and cotangent functions. It is also obtained from the Pythagorean theorem, but this time by dividing the formula by the square of the opposite side. The identity demonstrates that the square of the cosecant of an angle equals one plus the square of the cotangent of the same angle. It assists in representing one trigonometric function in terms of another and finds applications in numerous trigonometric simplifications. The identity can mathematically be written as:
The identity follows the rule for all values of A as “0°＜A90°”
The trigonometric identities help simplify the process of writing different trigonometric ratios in terms of another ratio. For example, different ratios of trigonometry can be written in terms of “Sin” by interchanging and doing arithmetic operations on the three identities mentioned earlier. This forms a basic concept for solving different questions of exercise 8.4.
Mastering the three trigonometric identities will not only sharpen your problem-solving skills but also lay a strong foundation for your future in mathematics. This is why you should keep practicing these identities with NCERT Solutions Class 10 Maths Chapter 8 Exercise 8.4.
1. Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.
Solution:
We have cosec² A - cot² A = 1
=> cosec² A = 1 + cot² A
=> (cosec A)² = cot² A + 1
=> (1/sin A)² = cot² A + 1
=> (sin A)² = 1/(cot² A + 1)
=> sin A = ± 1/√(cot² A + 1)
We reject the negative value of sin A for acute angle A. Therefore, sin A = 1/√(cot² A + 1).
Also, tan A = 1/cot A.
We have sec² A - tan² A = 1
=> sec² A = 1 + tan² A
=> sec² A = 1 + 1/cot² A = (cot² A + 1)/cot² A
=> sec A = √(cot² A + 1)/cot A
2. Write all the other trigonometric ratios of ∠A in terms of sec A.
Solution:
(i) sin A = √(1 - cos² A)
= √(1 - 1/sec² A) = √(sec² A - 1)/sec A
(ii) cos A = 1/sec A
(iii) tan A = √(sec² A - 1)
(iv) cot A = 1/tan A = 1/√(sec² A - 1)
(v) cosec A = 1/sin A = sec A/√(sec² A - 1)
3. Evaluate:
(i) (sin² 63° + sin² 27°) / (cos² 17° + cos² 73°)
(ii) sin 25° cos 65° + cos 25° sin 65°
Solution:
(i) (sin² 63° + sin² 27°) / (cos² 17° + cos² 73°)
= {sin(90° - 27°)}² + sin² 27° / cos² 17° + {cos(90° - 17°)}²
= {cos 27°}² + sin² 27° / cos² 17° + {sin 17°}²
= (cos² 27° + sin² 27°) / (cos² 17° + sin² 17°) = 1/1 = 1
(ii) sin 25° cos 65° + cos 25° sin 65°
= sin(90° - 65°) cos 65° + cos(90° - 65°) sin 65°
= cos 65° cos 65° + sin 65° sin 65°
= cos² 65° + sin² 65° = 1
4. Choose the correct option. Justify your choice:
(i) 9 sec² A - 9 tan² A =
(A) 1
(B) 9
(C) 8
(D) 0
(ii) (1 + tan θ + sec θ)(1 + cot θ - cosec θ) =
(A) 0
(B) 1
(C) 2
(D) -1
(iii) (sec A + tan A)(1 - sin A) =
(A) sec A
(B) sin A
(C) cosec A
(D) cos A
(iv) (1 + tan² A) / (1 + cot² A) =
(A) sec² A
(B) -1
(C) cot² A
(D) tan² A
Solution:
(i) Option (B)
9 sec² A - 9 tan² A = 9(sec² A - tan² A) = 9 × 1 = 9.
(ii) Option (C)
(1 + tan θ + sec θ)(1 + cot θ - cosec θ)
= {1 + sin θ/cos θ + 1/cos θ} × {1 + cos θ/sin θ - 1/sin θ}
= {(cos θ + sin θ + 1)/cos θ} × {(sin θ + cos θ - 1)/sin θ}
= {(cos θ + sin θ) + 1} × {(cos θ + sin θ) - 1} / (cos θ × sin θ)
= {(cos θ + sin θ)² - 1²} / (cos θ × sin θ)
{∵ (a + b)(a - b) = a² - b²}
= (cos² θ + sin² θ + 2 cos θ sin θ - 1) / (cos θ × sin θ)
= (1 + 2 cos θ sin θ - 1) / (cos θ sin θ) = 2.
(iii) Correct option is (D).
(sec A + tan A)(1 - sin A)
= sec A - sec A sin A + tan A - tan A sin A
= 1/cos A - sin A/cos A + sin A/cos A - sin² A/cos A
= (1 - sin² A) / cos A
= cos² A / cos A = cos A
(iv) Correct option is (D).
(1 + tan² A) / (1 + cot² A) = sec² A / cosec² A = tan² A
5. Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
(i) (cosec θ - cot θ)² = (1 - cos θ) / (1 + cos θ)
(ii) (cos A / (1 + sin A)) + ((1 + sin A) / cos A) = 2 sec A
(iii) (tan θ / (1 - cot θ)) + (cot θ / (1 - tan θ)) = 1 + sec θ cosec θ
(iv) (1 + sec A) / sec A = (sin² A) / (1 - cos A)
(v) (cos A - sin A + 1) / (cos A + sin A - 1) = cosec A + cot A, using the identity cosec² A = 1 + cot² A.
(vi) √((1 + sin A) / (1 - sin A)) = sec A + tan A
(vii) (sin θ - 2 sin³ θ) / (2 cos³ θ - cos θ) = tan θ
(viii) (sin A + cosec A)² + (cos A + sec A)² = 7 + tan² A + cot² A
(ix) (cosec A - sin A)(sec A - cos A) = 1 / (tan A + cot A)
(x) ((1 + tan² A) / (1 + cot² A)) = ((1 - tan A) / (1 - cot A))² = tan² A
Solution:
(i) LHS = (cosec θ - cot θ)²
= (1/sin θ - cos θ/sin θ)² = ((1 - cos θ) / sin θ)²
= (1 - cos θ)² / sin² θ = (1 - cos θ)² / (1 - cos² θ)
= (1 - cos θ)² / ((1 - cos θ)(1 + cos θ)) = (1 - cos θ) / (1 + cos θ)
∴ LHS = RHS.
(ii) LHS = (cos A / (1 + sin A)) + ((1 + sin A) / cos A)
= (cos² A + (1 + sin A)²) / (cos A(1 + sin A))
= (cos² A + 1 + sin² A + 2 sin A) / (cos A(1 + sin A))
= (2(1 + sin A)) / (cos A(1 + sin A))
= 2 sec A = RHS.
(iii) LHS = (tan θ / (1 - cot θ)) + (cot θ / (1 - tan θ))
= ((sin θ / cos θ) / (1 - cos θ / sin θ)) + ((cos θ / sin θ) / (1 - sin θ / cos θ))
= ((sin θ / cos θ) / (1 - cos θ / sin θ)) + ((cos θ / sin θ) / (1 - sin θ / cos θ))
= (sin θ × sin θ) / (cos θ(sin θ - cos θ)) + (cos θ × cos θ) / (sin θ(cos θ - sin θ))
= (sin² θ) / (cos θ(sin θ - cos θ)) - (cos² θ) / (sin θ(sin θ - cos θ))
= (sin³ θ - cos³ θ) / (cos θ sin θ (sin θ - cos θ))
= ((sin θ - cos θ)(sin² θ + cos² θ + sin θ cos θ)) / (cos θ sin θ (sin θ - cos θ))
{∵ a³ - b³ = (a - b)(a² + b² + ab)}
= (sin² θ + cos² θ + sin θ cos θ) / (cos θ sin θ)
= (1 + sin θ cos θ) / (cos θ sin θ) = 1 / (cos θ sin θ) + 1
= 1 + (1 / cos θ)(1 / sin θ)
= 1 + sec θ cosec θ
∴ LHS = RHS.
(iv) LHS = (1 + sec A) / sec A = (1 + 1/cos A) / (1/cos A) = (cos A + 1) / 1 = 1 + cos A
RHS = (sin² A) / (1 - cos A) = (1 - cos² A) / (1 - cos A) = 1 + cos A
∴ LHS = RHS.
(v) LHS = (cos A - sin A + 1) / (cos A + sin A - 1)
= ((cos A / sin A) - (sin A / sin A) + (1 / sin A)) / ((cos A / sin A) + (sin A / sin A) - (1 / sin A))
(Dividing the numerator and denominator by sin A)
= (cot A - 1 + cosec A) / (cot A + 1 - cosec A) = ((cosec A + cot A) - 1) / (1 + cot A - cosec A)
= ((cosec A + cot A) - (cosec² A - cot² A)) / (1 + cot A - cosec A)
(∵ cosec² A = 1 + cot² A, i.e., cosec² A - cot² A = 1)
= ((cosec A + cot A) - (cosec A + cot A)(cosec A - cot A)) / (1 + cot A - cosec A)
{∵ (a + b)(a - b) = a² - b²}
= ((cosec A + cot A)(1 - (cosec A - cot A))) / (1 + cot A - cosec A)
= ((cosec A + cot A)(1 + cot A - cosec A)) / (1 + cot A - cosec A)
= cosec A + cot A = RHS.
(vi) LHS = √((1 + sin A) / (1 - sin A)) = √(((1 + sin A)(1 + sin A)) / ((1 - sin A)(1 + sin A)))
= √((1 + sin A)² / (1 - sin² A)) = √((1 + sin A)² / cos² A) = (1 + sin A) / cos A
= 1 / cos A + sin A / cos A = sec A + tan A
∴ LHS = RHS.
(vii) LHS = (sin θ - 2 sin³ θ) / (2 cos³ θ - cos θ)
= (sin θ(1 - 2 sin² θ)) / (cos θ(2 cos² θ - 1))
= (sin θ(sin² θ + cos² θ - 2 sin² θ)) / (cos θ(2 cos² θ - sin² θ - cos² θ))
= (tan θ(cos² θ - sin² θ)) / (cos² θ - sin² θ)
= tan θ = RHS.
(viii) LHS = (sin A + cosec A)² + (cos A + sec A)²
= sin² A + cosec² A + 2 + cos² A + sec² A + 2
= 4 + 1 + 1 + cot² A + 1 + tan² A
= 7 + tan² A + cot² A = RHS.
(ix) LHS = (cosec A - sin A)(sec A - cos A)
= (1/sin A - sin A)(1/cos A - cos A)
= ((1 - sin² A) / sin A)((1 - cos² A) / cos A)
= (cos² A / sin A)(sin² A / cos A) = sin A cos A
RHS = 1 / (tan A + cot A) = 1 / ((sin A / cos A) + (cos A / sin A))
= 1 / ((sin² A + cos² A) / (sin A cos A)) = (sin A cos A) / 1 = sin A cos A
∴ LHS = RHS.
(x) LHS = (1 + tan² A) / (1 + cot² A) = sec² A / cosec² A = tan² A = RHS.
((1 - tan A) / (1 - cot A))² = (((cos A - sin A) / cos A) / ((sin A - cos A) / sin A))²
= (sin² A / cos² A) = tan² A = RHS.
(Session 2025 - 26)