Introduction to Trigonometry
Question 1.
Evaluate :
(i) sin18∘/cos72∘
(ii) tan26∘/cot64∘
(iii) cos 48° – sin 42°
(iv) cosec 31° – sec 59°
Solution:
(i) sin18∘/cos72∘
= sin(90∘−72∘)/cos72∘
= cos72∘/cos72∘
= 1
(ii) tan26∘/cot64∘
= tan(90∘−64∘)/cot64∘
= cot64∘/cot64∘
= 1
(iii) cos 48° – sin 42°
= cos (90° – 42°) – sin 42°
= sin 42° – sin 42° = 0
(iv) cosec 31° – sec 59°
= cosec (90° – 59°) – sec 59°
= sec 59° – sec 59° = 0
Question 2.
Show that
(i) tan 48° tan 23° tan 42° tan 67° = 1
(ii) cos 38° cos 52° – sin 38° sin 52° = 0
Solution:
(i) tan 48° tan 23° tan 42° tan 67°
= tan (90° – 42°) tan (90° – 67°) tan 42° tan 67°
= cot 42° cot 67° tan 42° tan 67°
= 1/tan42∘ x 1/tan67∘ x tan 42° x tan 67°
= 1
(ii) cos 38° cos 52° – sin 38° sin 52°
= cos (90° – 52°) cos (90° – 38°) – sin 38° sin 52°
= sin 52° sin 38° – sin 38° sin 52°
= 0
Question 3.
If tan 2A = cot (A – 18°), where 2A is an acute angle, find the value of A.
Solution:
We are given that
tan 2A = cot (A – 18°) … (1)
Since tan 2A = cot (90° – 2A), so we can write (1) as
cot (90° – 2A) = cot (A – 18°)
Since (90° – 2A) and (A – 18°) are both acute angles, therefore
90° – 2A = A – 18°
or – 2A – A = – 18° – 90°
or – 3A = – 108°
or A = 36°
Question 4.
If tan A = cot B, prove that A + B = 90°.
Solution:
We are given that
tan A = cot B … (1)
Since tan A = cot (90° – A), so we can write (1)
as cot (90° – A) = cot B
Since (90° – A) and B are both acute angles, therefore
90° – A = B
or A + B = 90°
Question 5.
If sec 4A = cosec (A – 20°), where 4A is an acute angle, find the value of A.
Solution:
We are given that
sec 4A = cosec (A – 20°). …(1)
Since sec 4A = cosec (90° – 4A), so we can write (1) as
cosec (90° – 4A) = cosec (A – 20°)
Since (90° – 4A) and (A – 20°) are both acute angles, therefore
90° – 4A = A – 20°
or – 4A – A = – 20° – 90°
or – 5A = – 110°
or A = 22°
Question 6.
If A, B and C are, interior angles of a ∆ ABC, then show, that
sin((B+C)/2) = cosA/2
Solution:
Since A, B and C are the interior angles of a ∆ ABC, therefore
Question 7.
Express sin 67° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°.
Solution:
sin 67° + cos 75°
= sin (90° – 23°) + cos (90° – 15°)
= cos 23° + sin 15°
