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(i) cos²θ + cosθ = 1 

LHS = cos² θ + cos θ 

=1 − sin² θ + cos θ 

= 1 − (sin² θ − cos θ) 

Since LHS ≠ RHS, this not an identity.

(ii) sin²θ + sinθ =  1

LHS = sin² θ + sin θ 

= 1 − cos² θ + sin θ

= 1 − (cos² θ − sin θ) 

Since LHS ≠ RHS, this is not an identity.

(iii) tan²θ + sinθ = cos²θ

LHS = tan² θ + sin θ

= \(\frac{sin^2θ}{cos^2θ}\) + sinθ

= \(\frac{1-cos^2θ}{cos^2θ}\)+ sinθ

= sec² θ − 1 + sin θ 

Since LHS ≠ RHS, this is not an identity.

cot (– 870°) = – cot 870°               (∵ cot (– θ) = – cot θ) 

= – cot (720° + 150°) = – cot (2 x 360° + 150°)

= – cot 150° = – cot (90° + 60°)          (∵ cot (n .360° + θ) = cot θ, n ∈ N) 

= – (– tan 60°)                          (∵cot (90° + θ) = – tan θ) 

= tan 60° = √3 .

Answer : (a) \(\frac{nπ}{2}+\frac{π}{8}\), n ∈ I

(sin 3x + sin x) – 3 sin 2x = (cos 3x + cos x) – 3 cos 2x 

⇒ 2 sin 2x cos x – 3 sin 2x = 2 cos 2x . cos x – 3 cos 2x 

⇒ sin 2x (2 cos x – 3) = cos 2x (2 cos x – 3) 

⇒ (2 cos x – 3) (sin 2x – cos 2x) = 0 

⇒ (2 cos x – 3) = 0 or (sin 2x – cos 2x) = 0 

⇒ cos x = \(\frac{3}{2}\) or sin 2x = cos 2x 

∵  – 1 ≤ cos x ≤ 1, cos x ≠  \(\frac{3}{2}\)

∴ sin 2x = cos 2x ⇒ tan 2x = 1 

⇒ tan 2x = tan (π/4) 

⇒ 2x = nπ + π/4

⇒ x = \(\frac{nπ}{2}+\frac{π}{8}\), n ∈ I

We know that

cosec (30°) = 2

Tan(45°) = 1

sec (60°) = 2

Now putting the values;

= (2)² × (1)² - (2)²

= 4 – 4

= 0

We know that

cosec (30^o) = 2

Tan(45^o) = 1

sec (60^o) = 2

Now putting the values;

= (2)² × (1)² - (2)²

= 4 – 4

= 0

We know that

Sin (90^o) = 1

Cos (0^o) = 1

Tan(0^o) = 0

Tan(45^o) = 1

Now putting the value, we get

= 1 – 1 + 0 + 1

= 1

tan² 30° + tan² 60° + tan² 45°

= \((\frac{1}{\sqrt3})^2\) + \((\sqrt3)^2\) + (1)²

^{= \(\frac{1}3+3+1\) = \(\frac{13}3\)}

cos² 30° + cos² 45° + cos² 60°+ cos² 90°

= \((\frac{\sqrt3}{2})^2\) + \((\frac{1}{\sqrt{2}})^2\) = \((\frac{1}{2})^2\) + (0)²

^{= \(\frac{3}4\) + \(\frac{1}2\) + \(\frac{1}4\) = \(\frac{3}2\)}

sin² 30° + sin² 45° + sin² 60°+ sin² 90°

= \((\frac{1}2)^2\) + \((\frac{1}{\sqrt2})^2\) + \((\frac{\sqrt3}2)^2\) + (1)² 

^{= \(\frac{1}4\) + \(\frac{1}2\) + \(\frac{3}4\) + 1 = \(\frac{5}2\)}

cos 0° = 1 cos 30° = \(\frac{\sqrt3}{2}\) = 0.87 

cos 45° = \(\frac{1}{\sqrt 2}\)= 0.7 cos 60° = \(\frac{1}{2}\)= 0.5 

cos 90° = 1 

∴ The value of cos 0 increases as 0 increases – the statement is False.

False. 

sin 30° = \(\frac{1}{2}\) cos 30° = \(\frac{\sqrt 3}{2}\)

sin 30° ≠ cos 30° 

But sin 45° = cos 45° = \(\frac{1}{\sqrt 2}\)

Taking LHS = (1 – cosθ)(1+ cosθ)

Using identity , (a + b) (a – b) = (a² – b²) , we get

= (1)² – (cosθ)²

= 1 – cos² θ

= sin² θ [∵ cos² θ + sin² θ = 1]

= RHS

Hence Proved

(c) √2

sec (90 – θ)° sin θ sec 45° 

= cosec θ sin θ. (√2) = \(\frac1{sin\,\theta}\) . sin θ = √2 = √2.

Taking LHS

= sin 42° cos 48° + cos 42° sin 48°

= cos (90° - 42°) cos 48° + sin (90° - 42°) sin 48°

[∵ Sin θ = cos (90° - θ) and cos θ = sin (90° - θ) ]

= cos 48° cos 48° + sin 48° sin 48°

= cos² 48° + sin² 48°

= 1 [∵ cos² θ + sin² θ = 1]

= LHS = RHS

Hence Proved

(a) \(\frac{-1}{\sqrt{2}}\)

\(\frac{1}{cosec(-45°)}\) = sin(-45°)

= -sin 45° 

= \(\frac{-1}{\sqrt{2}}\)

(c) 0

cos² 45° – sin² 45° 

= cos 2 x 45° (∵ cos² A – sin² A = cos 2A) 

= cos 90° 

= 0

(d) 0

1 – 2 sin² 45° 

= cos(2 x 45°) [∵ cos 2A = 1 – 2 sin² A] 

= cos 90° 

= 0

Taking LHS = cos θ cos ( 90° - θ) + sin θ sin (90° - θ)

⇒ cos θ × sin θ – sinθ × cos θ [∵ Sin θ = cos (90° - θ) and cos θ = sin (90° - θ) ]

= 0 = RHS

Hence Proved

tan1°tan2°......tan89°

= tan 1° tan 2° tan 3° … tan 45° … tan 87° tan 88° tan 89° 

= tan 1° tan 2° tan 3° … tan 45° … cot(90° − 87°) cot(90° − 88°) cot(90° − 89°) 

= tan 1° tan 2° tan 3° … tan 45° … cot 3° cot 2° cot 1° 

= tan 1° × tan 2° × tan 3° × …×1× …× 1/tan 3° × 1/tan 2° × 1/tan 1° 

= 1

Taking LHS = sin θ cos (90° - θ) + cos θ sin (90° - θ)

⇒ sin θ × sin θ + cos θ × cos θ [∵ Sin θ = cos (90° - θ) and cos θ = sin (90° - θ) ]

⇒ cos² θ + sin² θ [∵ cos² θ + sin² θ = 1]

= 1 = RHS

Hence Proved

tan10° tan 20° tan 70° tan 80°

= cot(90° − 10° ) cot(90° − 20° ) tan70° tan80° 

= cot80° cot70° tan70° tan80°

= 1/tan 80° × 1/tan 70° × tan 70° × tan 80° 

= 1

Taking LHS= sin²29^o + sin²61^o

⇒ cos² (90° - 29°) + sin² 61° [∵ Sin θ = cos (90° - θ)]

⇒ cos² 61° + sin² 61°

= 1 =RHS [∵ cos² θ + sin² θ = 1]

Hence Proved

(b) sin 50°

p sec 50° = tan 50°

p(\(\frac{1}{cos50°}\)) = \(\frac{sin50°}{cos50°}\)

∴ p = sin 50°