Trigonometry Exercise 13.4 – Solutions:
- Evaluate:
- tan65˚/cot25˚
- sin18˚/cos72˚
iii. cos48˚- sin42˚
- cosec31˚ – sec59˚
- cot34˚ – tan56˚
- sin36˚/cos54˚ – sin54˚/cos36˚
vii. sec70˚ sin20˚ – cos70˚cosec20˚
viii. cos213˚ – sin277˚
- Prove that:
- sin35˚ sin55˚ – cos35˚cos55˚ = 0
- tan10˚tan15˚tan75˚tan80˚ = 1
iii. cos38˚cos52˚ – sin38˚sin52˚ = 0
III. If sin5θ = cos4θ, where 4θ and 5θ are acute angles, find the value of θ
- If sec4A = cosec(A – 20˚), where 4A is an acute angle, find the value of A.
Trigonometry Exercise 13.4 – Questions:
I. Evaluate:
- tan65˚/cot25˚
Solution:
We know,
cot25˚ = tan(90˚ – 25˚) = tan65˚
Therefore, tan65˚/cot25˚ = tan65˚/tan65˚ = 1
ii. sin18˚/cos72˚
Solution:
We know,
cos72˚ = sin(90˚ – 72˚) = sin18˚
Therefore, sin18˚/cos72˚ = sin18˚/sin18˚ = 1
iii. cos48˚- sin42˚
Solution:
We know,
cos48˚ = sin(90˚ – 48˚) = sin48˚
Therefore, cos48˚- sin42˚= cos48˚- cos48˚ = 0
iv. cosec31˚ – sec59˚
Solution:
We know,
cosec31˚ = sec(90˚ – 31˚) = sec59˚
Therefore, cosec31˚ – sec59˚= cosec31˚ – cosec31˚= 0
v. cot34˚ – tan56˚
Solution:
We know,
cot34˚ = tan(90˚ – 34˚) = tan56˚
Therefore, cot34˚ – tan56˚= cot34˚ – cot34˚= 0
vi. sin36˚/cos54˚ – sin54˚/cos36˚
Solution:
We know,
cos54˚ = sin(90˚ – 54˚) = sin36˚
cos36˚ = sin(90˚ – 36˚) = sin54˚
Therefore, sin36˚/cos54˚ – sin54˚/cos36˚ = sin36˚/sin36˚ – sin54˚/sin54˚= 1 – 1 = 0
vii. sec70˚ sin20˚ – cos70˚cosec20˚
Solution:
We know,
sec70˚ = cosec(90˚ – 70˚) = cosec20˚
sec20˚ = cos(90˚ – 20˚) = cos70˚
Therefore, sec70˚ sin20˚ – cos70˚cosec20˚ = cosec20˚ cos70˚ – cos70˚cosec20˚= 0
viii. cos213˚ – sin277˚
Solution:
We know, cos213˚ – sin277˚
cos213˚ = 1 – sin213˚
sin213˚ = cos2(90˚ – 13˚) = sin2(77˚)
Therefore, cos213˚ – sin277˚ = 1 – sin277˚- sin277˚= 1
II. Prove that:
- sin35˚ sin55˚ – cos35˚cos55˚ = 0
Solution:
sin35˚ sin55˚ – cos35˚cos55˚ = 0
cos35˚ = sin(90˚ – 35˚) = sin55˚
cos55˚ = sin(90˚ – 55˚) = sin35˚
Therefore, sin35˚ sin55˚ – cos35˚cos55˚
= sin35˚ sin55˚ – sin35˚ sin55˚
= 0
ii. tan10˚tan15˚tan75˚tan80˚ = 1
Solution:
⇒ sin10˚/cos10 . sin15˚/cos15. sin75˚/cos75. sin80˚/cos80 = 1
cos 10˚ = sin(90˚ – 10˚) = sin 80˚
cos 15˚ = sin(90˚ – 15˚) = sin75˚
cos75˚ = sin(90˚ – 75˚) = sin15˚
cos80˚ = sin(90˚ – 10˚) = sin 10˚
⇒ sin10˚/sin80˚ . sin15˚/sin75˚. sin75˚/sin15˚. sin80˚/sin10˚ = 1
⇒ tan10˚tan15˚tan75˚tan80˚ = 1
iii. cos38˚cos52˚ – sin38˚sin52˚ = 0
Solution:
cos38˚cos52˚ – sin38˚sin52˚ = 0
we know,
cos38 = sin(90 – 38) = sin 52
cos52 = sin(90 – 52) = sin38
⇒ cos38˚cos52˚ – sin38˚sin52˚ = sin52˚sin38˚ – sin38˚sin52˚ = 0
III. If sin5θ = cos4θ, where 4θ and 5θ are acute angles, find the value of θ
Solution:
Given, sin5θ = cos4θ
We know, cos4θ = sin(90˚ – 4θ)
Since, sin(90˚ – 4θ) = sin5θ
⇒ 90˚ – 4θ = 5θ
⇒ 90˚ = 9θ
⇒ θ = 90˚/9 = 10˚
IV. If sec4A = cosec(A – 20˚), where 4A is an acute angle, find the value of A.
Solution:
Given sec4A = cosec(A – 20˚)
sec4A = cosec(90˚ – 4A)
Since cosec(90˚ – 4A) = cosec(A – 20˚)
90˚ – 4A = A – 20˚
90˚ + 20˚ = A + 4A
110˚ = 5A
A = 110˚/5 = 22˚