Introduction to Trigonometry
Mathematics Class Tenth
NCERT Exercise 8.4 solution part3
Question (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
Solution:
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)2/12 – cos2 θ
= (1 – cos θ) (1 – cos θ)/(1 + cos θ) (1 – cos θ)
= 1 – cos θ/1 + cos θ Proved
(ii) cos A/1 + sin A + 1 + sin A/cos A = 2 sec A
Solution:
Given, LHS = cos A/1 + sin A + 1 + sin A/cos A
= cos2A + (1 + sin A)2/(1+sin A) cos A
=cos2 A + 12 + sin2A + 2sinA/(1 + sin A) cos A
= (cos2A + sin2A) + 1 + 2sin A/(1+sin A) cos A
[∵ cos2A + sin2A = 1]
= 1 + 1 + 2sin A/(1+sin A) cos A
= 2 + 2sin A/(1+sin A) cos A
By taking 2 as common from numerator
= 2 (1 + sin A)/(1 + sin A) cos A
= 2/cos A
= 2 × 1/cos A
= 2 sec A = RHS Proved
(iii) tan θ/1 – cot θ + cot θ/1 – tan θ = 1 + sec θ cosec θ
[Hint : Write the expression in terms of sin θ and cos θ]
Solution:
Given, 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/sin θ – cos θ/sin θ + cos θ/sin θ/cos θ – sin θ/cos θ
= sin θ/cos θ × sin θ/sin θ – cos θ + cos θ/sin θ × cos θ/cos θ – sin θ
= sin2 θ/cos θ (sin θ – cos θ) + cos2 θ/sinθ (cos θ – sin θ)
Taking minus (–) as common in the denominator of the second term of the above expression
= sin2 θ/cos θ (sin θ – cos θ) + cos2 θ/sinθ (– sin θ + cos θ)
= sin2 θ/cos θ (sin θ – cos θ) + cos2 θ/– sinθ (sin θ – cos θ)
= sin2 θ/cos θ (sin θ – cos θ) – cos2 θ/ sinθ (sin θ – cos θ)
Now after taking 1/sin θ – cos θ as common
= 1/sinθ – cosθ (sin2θ/cosθ – cos2θ/sinθ)
= 1/sinθ – cosθ (sin3 θ – cos3 θ/sin θ cos θ)
= 1/sinθ – cosθ [(sin θ – cos θ) (sin2θ + cos2 θ + sin θ cos θ)/sin θ cos θ]
= sin2θ + cos2 θ + sin θ cos θ/sin θ cos θ
= 1 + sin θ cos θ/sin θ cos θ
[∵ sin2 θ + cos2 θ = 1]
= 1/sin θ cos θ + sin θ cos θ/sin θ cos θ
= 1/sin θ cos θ + 1
= 1/sin θ × 1/cos θ + 1
= cosec θ sec θ + 1
= sec θ cosec θ + 1
= 1 + sec θ cosec θ Proved
(iv) 1 + sec A/sec A = sin2 A/1 – cos A
Solution:
Given, LHS = 1 + sec A/sec A
= 1 + 1/cos A/1/cos A
= cos A + 1/cos A/1/cos A
= cos A + 1/cos A × cos A/1
= cos A + 1
= 1 + cos A
After mulitplying with = 1 – cos A/1 – cos A
= 1 + cos A × 1 – cos A/1 – cos A
= (1 + cos A) (1 – cos A)/1 – cos A
= 12 – cos2 A/1 – cos A
= 1 – cos2 A/1 – cos A
= sin2 A/1 – cos A Proved
(v) cos A – sin A + 1/cos A + sin A – 1 = cosec A + cot A
Solution:
Given, LHS = cos A – sin A + 1/cos A + sin A – 1
After dividing numerator and denominator by sin A we get
= cos A – sin A + 1/sinA/cos A + sinA –1/sin A
= cos A/sin A – sin A/sin A + 1/sin A/cos A/sin A + sin A/sin A – 1/sin A
= cot A – 1 + cosec A/cot A + 1 – cosec A
= cot A – (1 – cosec A)/cot A + (1 – cosec A)
After multiplying with cot A – (1 – cosec A)/cot A – (1 – cosec A) we get
= 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 – cosec A)2
= cot2 A + (1 –cosec A)2 – 2 cot A (1 – cosec A)/cot2 A – (12 + cosec2 A – 2 cosec A)
= cot2 A + 12 + cosec2 A – 2 cosec A – 2 cot A + 2 cot A cosec A/cot2 A – 1 – cosec2 A + 2 cosec A
= (cot2 A + 1) + cosec2 A – 2 cosec A – 2 cot A + 2 cot A cosec A/(cot2 A – cosec2 A) – 1 + 2 cosec A
= cosec2 A + cosec2 A – 2 cosec A – 2 cot A + 2 cot A cosec A/– 1 – 1 + 2 cosec A
[∵ (cot2 A + 1 = cosec2 A) and (cot2 A – cosec2 A = –1)]
= 2cosec2 A + 2 cot A cosec A – 2 cosec A – 2 cot A/2 cosec A – 2
= 2 cosec A (cosec A + cot A) – 2 (cosec A + cot A)/2 cosec A – 2
After taking (cosec A + cot A) as common
= (cosec A + cot A) (2 cosec A – 2)/2 cosec A – 2
= cosec A + cot A Proved
Reference: