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Prove the following identities, where the angles involved are acute angles for which the expressions are defined
(i) (cosecθ - cotθ)² = (1 - cosθ)/(1 + cosθ)
(ii) cos A/(1 + sin A) + (1 + sin A)/cos A = 2 sec A
(iii) tanθ/(1 - cotθ) + cotθ/(1 - tanθ) = 1 + secθ cosecθ [Hint : Write the expression in terms of sin θ and cos θ]
(iv) (1 + sec A)/sec A = sin² A/(1 - cos A) [Hint :Simplify LHS and RHS separately]
(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) (sinθ - 2sin³θ)/(2cos³θ - cosθ) = tanθ
(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) [Hint : Simplify LHS and RHS separately]
(x) (1 + tan² A)/(1 + cot² A) = (1 - tan A / 1 - cot A)² = tan² A

Solution:

We use the standard identities of trigonometric and basic properties of trigonometric ratios to solve the given problems.

sin² A + cos² A = 1

cosec² A = 1 + cot² A

sec² A = 1 + tan² A

(i) (cosecθ - cotθ)² = (1 - cosθ) / (1 + cosθ)

L.H.S = (cosecθ - cotθ)² .....(1)

We know that the trigonometric functions,

cot (x) = cos (x)/sin (x) = 1/tan (x)

cosec (x) = 1/sin (x)

By substituting the above function in Equation (1)

(cosecθ - cot θ)² = (1/sinθ - cosθ/sinθ)²

= (1 - cosθ)²/(sinθ)²

= (1 - cosθ)²/sin² θ

= (1 - cosθ)²/(1 - cos²θ) (By Identity sin A + cos A = 1 Hence, 1 - cos A= sin A)

= (1 - cosθ)² (1 - cosθ)(1 + cosθ) [Using a² - b² = (a + b)(a - b)]

= (1 - cosθ)/(1 + cosθ)

= RHS

(ii) cos A / (1 + sin A) + (1 + sin A) / cos A = 2 sec A

L.H.S = cosA / (1 + sinA) + (1 + sinA) / cosA

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

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

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

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

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

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

= 2/cos A

= 2sec A

= R.H.S

(iii) tanθ/(1 - cotθ) + cotθ/(1 - tanθ) = 1 + secθ cosecθ

LHS = tanθ/(1 - cotθ) + cotθ/(1 - tanθ)

We know that the trigonometric functions,

tan (x) = sin (x)/cos (x)

cot (x) = cos (x)/sin (x) = 1/tan (x)

By substituting the above relations in Equation (1),

= [(sinθ/cosθ)/(1 - cosθ/sinθ)] + [(cosθ/sin θ)/1 - cosθ/sinθ]

= [(sinθ/cosθ)(sinθ - cosθ)/sinθ] + [(cosθ/sin θ)/(cosθ - sin θ)/cosθ

= sin² θ/cosθ(sinθ - cosθ) + cos² θ/sin θ(sinθ - cosθ)

Taking 1/(sinθ - cosθ) as common

= 1/(sinθ - cosθ) [sin²θ/cosθ - cos²θ/sinθ]

= 1/(sinθ - cosθ)[(sin³θ - cos³θ)/sinθ cosθ]

Using a³ - b³ = (a - b)(a² + ab + b²)

= 1/(sinθ - cosθ) [((sinθ - cosθ) sin²θ+ cos²θ + sin θcosθ)/sinθ cosθ]

= (1 + sinθ cosθ)/(sinθ cosθ) (By Identity sin² A+ cos² A = 1)

= 1/sinθ cosθ + (sinθ cosθ/sinθ cosθ)

= 1 + secθ cosecθ

= R.H.S.

(iv) (1 + sec A)/sec A = sin² A/(1 - cos A)

L.H.S = (1 + sec A)/sec A ......(1)

We know that the trigonometric functions,

sec (x) = 1/cos (x)

By substituting the above function in Equation (1),

(1 + sec A)/sec A = (1 + 1/cos A)/(1/cos A)

= (cos A + 1)/cosA/(1/cos A)

= (cosA + 1)/cos A × cos A/1

= (1 + cos A)

By multiplying (1- cos A), in both denominator and numerator

⇒ [(1 - cos A)(1 + cos A)] / (1 - cos A)

= (1 - cos² A)/1 - cos A [By Identity sin A + cos A = 1]

= sin² A/1 - cos A

= R. H.S

(v) (cos A - sin A + 1)/cos A + sin A + 1 = cosec A + cot A

L.H.S = cos A - sin A + 1/(cos A + sin A - 1)

Diving both numerator and denominator by sin A

= [cos A/sin A - sin A/sin A + 1/sin A]/[cos A/sin A + sin A/sin A - 1/sin A]

We know that the trigonometric functions,

cot (x) = cos(x)/sin (x) = 1/tan (x)

cosec(x) = 1/sin (x)

We get,

⇒ (cot A - 1 + cosec A) / (cot A + 1 - cosec A)

⇒ cot A - (1 - cosec A) / cot A+ (1 - cosec A)

We know that, 1 + cot² A = Cosec² A

Hence multiplying [cot A - (1 - cosec A)] in numerator and denominator

= [(cot A) - (1 - cosec A) × (cot A) - (1 - cosec A)]/[(cot A) + (1 - cosec A)-× (cot A) - (1 - cosec A)]

= [cot A - (1 - cosec A)]²/[(cot A)² - (1 - cosecA)²]

= [cot² A + (1 -cosecA)² - 2cot A(1 - cosecA)]/[cot² A - (1 + cosec² A - 2cosec A)]

= (cot² A + 1 + cosec² A - 2cosec A - 2cot A + 2cot AcosecA)/(cot²A - (1 + cosec² A - 2cosec A))

= (2cosec² A+ 2cot A cosec A - 2cot A - 2cosecA)/(cot² A - 1 - cosec² A + 2cosecA)

= 2cosec A(cosec A+ cot A) - 2(cot A + cosec A)/(cot² A - cosec² A - 1 + 2cosec A)

= (cosec A + cot A)(2cosec A - 2)/(- 1 - 1 + 2cosec A)

= (cosec A + cot A)(2cosec A - 2)/(2cosec A - 2)

= cosec A + cot A

= R.H.S

(vi) √1 + sin A/1 - sin A = sec A+ tan A

LHS = 1 + sin A/(1 - sin A) .....(1)

Multiplying and dividing by (1 + sin A)

⇒ (1 + sin A)(1 + sin A/1 - sin A)(1 + sin A)

= (1 + sin A)²/(1 - sin² A) [a² - b² = (a - b)(a + b)]

= (1 + sinA)/1 - sin² A

= 1 + sin A/cos² A

= 1 + sin A/cos A

= 1/cos A + sin A/cos A

= sec A + tan A

= R.H.S

(vii) (sinθ - 2sin³ θ)/(2cos³ θ - cosθ) = tanθ

L.H.S = (sinθ - 2sin³θ)/(2cosθ - cosθ)

Taking Sin θ and Cos θ common in both numerator and denominator respectively.

sinθ (1 - 2 sin²θ)/cosθ (2cos²θ - 1)

By Identity sin² A + cos² A = 1 hence, cos² A = 1 - sin² A and substituting this in the above equation,

⇒ sinθ (1 - 2sin²θ)/cosθ {2(1 - sin²θ) - 1}

= sinθ (1 - 2sin²θ)/cosθ(2 - 2sin²θ - 1)

= sinθ (1 - 2sin²θ)/cosθ(1 - 2sin²θ)

= sinθ/cosθ

= tanθ

= R.H.S

(viii) (sin A + cosec A)² + (cos A + sec A)² = 7 + tan² A + cot² A

L.H.S =-(sin A + cosec A)² + (cos A + sec A)²

By using (a + b)² = a² + 2ab + b²

⇒ sin² A + cosec² A + 2sin A cosec A + cos² A + sec² A + 2cos A sec A

By rearranging and using sec A = 1/cos A and cosec A = 1/sin A

⇒ (sin² A + cos² A) + (cosec² A + sec² A)+ 2 sin A (1/sin A) +-2cos A (1/cos A)

Hence (sin² A + cos² A) = 1, cosec² A = (1 + cot² A) and (sec² A- tan² A) = 1

⇒ 1 + 1 + cot² A + 1 + tan² A + 2 + 2

= 7 + tan² A + cot² A

= R.H.S

(ix) (cosec A - sin A)(sec A - cos A) = 1/(tan A + cot A)

L.H.S = (cosec A - sin A)(sec A - cos A) .....Equation (1)

We know that the trigonometric functions,

sec (x) = 1/cos (x)

cosec (x) = 1/sin (x)

By substituting the above relations in Equation (1)

⇒ (1/sin A - sin A)(1/cos A - cos A)

= (1 - sin² A)/sin A × (1 - cos² A)/cos A

= cos² A sin² A/sin A cos A

= sin A cos A/1

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

= 1/sin² A + cos² A [Dividing numerator and denominator by (sin A cos A)]

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

= 1/[(sin A/cos A) + (cos A/sin A)]

= 1/(tan A + cot A)

= RHS

(x) (1 + tan² A)/(1 + cot² A) = (1 - tan A/1 - cot A)² = tan² A

Taking LHS, (1 + tan² A)/(1 + cot² A)

= sec² A/cosec² A

= (1/cos²A)/(sin²A)

= (1/cos² A) × (sin² A/1)

= tan² A

= RHS

Taking,-(1 - tan A/1 - cot A)²

= [(1 - tan A)/(1 - 1/tan A)]²

= [(1 - tan A)/(tan A - 1)/tan A]²

= ((1 - tan A) × tan A(tan A - 1))²

=(-tanA)²

= tan² A

= R.H.S

Hence, L.H.S = R.H.S.

☛ Check: NCERT Solutions for Class 10 Maths Chapter 8

Video Solution:

Prove the following identities, where the angles involved are acute angles for which the expressions are defined(i) (cosecθ - cotθ)² = (1 - cosθ)/(1 + cosθ)(ii) cos A/(1 + sin A) + (1 + sin A)/cos A = 2 sec A(iii) tanθ/(1 - cotθ) + cotθ/(1 - tanθ) = 1 + secθ cosecθ [Hint : Write the expression in terms of sin θ and cos θ](iv) (1 + sec A)/sec A = sin² A/(1 - cos A) [Hint :Simplify LHS and RHS separately](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) (sinθ - 2sin³θ)/(2cos³θ - cosθ) = tanθ(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) [Hint : Simplify LHS and RHS separately](x) (1 + tan² A)/(1 + cot² A) = (1 - tan A / 1 - cot A)² = tan² A

Maths NCERT Solutions Class 10 Chapter 8 Exercise 8.4 Question 5

Summary:

We have proved the following identities, where the angles involved are acute angles for which the expressions are defined (i) (cosecθ - cotθ)² = (1 - cosθ)/(1 + cosθ) (ii) cos A/(1 + sin A) + (1 + sin A)/cos A = 2 sec A (iii) tanθ/(1 - cotθ) + cotθ/(1 - tanθ) = 1 + secθ cosecθ (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 (vi) √(1 + sin A/1 - sin A) = sec A+ tan A (vii) (sinθ - 2sin³ θ)/(2cos³ θ - cosθ) = tanθ (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

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