## Introduction to Trigonometry

Short Q and A

Q1: If sin 3A = cos (A – 26°), where 3A is an acute angle, find the value of A.

Answer: sin 3A = cos (A – 26°)

∵ sin 3A = cos (90° – 3A)
∴ cos (90° – 3A) = cos (A – 26°)

Since
90° – 3A and A – 26° are both acute angles, therefore,
90° – 3A = A – 26°
A = 29°

Q2: Express cot 85° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°.

Answer: cot 85° + cos 75° = cot (90° – 5°) + cos (90° – 15°)

= tan 5° + sin 15°

Q3: Evaluate cos 48° − sin 42°

Answer: cos 48° − sin 42°
= cos (90°− 42°) − sin 42°
= sin 42° − sin 42°
= 0

Q4: Show that:
(i) tan 48° tan 23° tan 42° tan 67° = 1
(ii) cos 38° cos 52° − sin 38° sin 52° = 0

(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°
= (cot 42° tan 42°) (cot 67° tan 67°)
= (1) (1)
= 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

Q5: If tan 2A = cot (A− 18°), where 2A is an acute angle, find the value of A.

Answer:  Given that, tan 2A = cot (A− 18°)
cot (90° − 2A) = cot (A −18°)
90° − 2A = A− 18°
108° = 3A
A = 36°

Q6: If tan A = cot B, prove that A + B = 90°

Answer: Given that, tan A = cot B
tan A = tan (90° − B)
A = 90° − B
A + B = 90°

Q7: If sec 4A = cosec (A− 20°), where 4A is an acute angle, find the value of A.

Answer: Given that, sec 4A = cosec (A − 20°)
cosec (90° − 4A) = cosec (A − 20°)
90° − 4A= A− 20°
110° = 5A
A = 22°

Q8: Express sin 67° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°.

Answer: sin 67° + cos 75°
= sin (90° − 23°) + cos (90° − 15°)
= cos 23° + sin 15°

Q9: Show that (1 - cos²θ)cosec²θ = 1

Answer: LHS = (1 - cos²θ)cosec²θ
∵ sin²θ +  cos²θ = 1
∴    = sin²θ cosec²θ
= sin²θ × 1/ sin²θ = 1

Q10: sec θ (1 - sin θ)(sec θ + tan θ) = 1

Answer: LHS = sec θ (1 - sin θ)(sec θ + tan θ)
= (sec θ − sec θ sin θ) (sec θ + tan θ)
= (sec θ − sin θ/cos θ) (sec θ + tan θ)
= (sec θ − tan θ) (sec θ + tan θ)
= sec²θ − tan²θ
= 1 = RHS

Q11: Show that sin⁶θ + cos⁶θ = 1 - 3sin²θcos²θ

Answer: LHS = sin⁶θ + cos⁶θ
= (sin²θ)³ + (cos²θ)³
= (sin²θ + cos²θ)(sin⁴θ + cos⁴θ - sin²θcos²θ)
= (1)[(sin²θ)² + (cos²θ)² + 2sin²θcos²θ - 3sin²θcos²θ]
= [(sin²θ + cos²θ)² - 3sin²θcos²θ]
= 1 - 3sin²θcos²θ = RHS

Q12: Prove that cosec(65° + θ) − sec(25° − θ) − tan(55° − θ) + cot(35° + θ) = 0

Answer: cosec(65° + θ) = cosec { 90° − (25° − θ)} = sec(25° − θ)
cot( 35° + θ ) = cot{ 90° − (55° − θ)} = tan(55° − θ)

∴ LHS = cosec(65° + θ) − sec(25° − θ) − tan(55° − θ) + cot(35° + θ)
= sec(25°− θ ) − sec( 25°− θ) − tan(55° − θ) + tan(55° − θ)
= 0 = RHS

CH 8: Introduction to Trigonometry (MCQs from CBSE Papers)
CH 8: Introduction to Trigonometry (NCERT Ex 8.1)
CH 8: Introduction to Trigonometry (NCERT Ex 8.2)
CH 8: Problems based on Trigonometric Identities
CH 8: Problems based on Trigonometric Identities (Set-2)
CH 8: Introduction to Trigonometry (Short Q & A)
CH 9: Applications of Trigonometry (MCQs) or Online Quiz