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Taking the L.H.S, 

= sin (50° + θ) – cos (40° – θ) + tan 1° tan 10° tan 20° tan 70° tan 80° tan 89° 

= sin (50° + θ) – sin (90° – (40° – θ)) + tan (90 – 89)° tan (90 – 80)° tan (90 – 70)° tan 70° tan 80° tan 89° [∵ sin (90 – θ) = cos θ] 

= sin (50° + θ) – sin (50° + θ) + cot 89° cot 80° cot 70° tan 70° tan 80° tan 89° [∵ tan (90° – θ) = cot θ]

= 0 + (cot 89° x tan 89°) (cot 80° x tan 80°) (cot 70° x tan 70°) 

= 0 + 1 x 1 x 1 [∵ tan θ x cot θ = 1] 

= 1

= R.H.S 

Hence Proved

Taking LHS = sin²40^o + sin²50^o

⇒ cos² (90° - 40°) + sin² 50° [∵ Sin θ = cos (90° - θ)]

⇒ cos² 50° + sin² 50°

= 1 =RHS [∵ cos² θ + sin² θ = 1]

Hence Proved

(i) LHS = sin (A + B) sin (A – B) 

= (sin A cos B + cos A sin B) (sin A cos B – cos A sin B) 

= sin² A cos² B – cos² A sin² B 

= sin² A (1 – sin² B) – (1 – sin² A) sin B 

= sin² A – sin² A sin² B – sin² B + sin² A sin² B 

= sin² A – sin² B = RHS

(ii) LHS = cos (A + B) cos (A – B) = (cos A cos B – sin A sin B) (cos A cos B + sin (A sin B) 

= cos² A cos² B – sin² A sin² B 

= cos² A (1 – sin² B) – (1 – cos² A) sin² B 

= cos² A – cos² A sin² B – sin² B + cos² A sin² B 

= cos² A – sin² B = RHS 

Now cos² A – sin² B = (1 – sin² A) – (1 – cos² B) 

= 1 – sin² A – 1 + cos² B 

= cos² B – sin² A

(iii) sin² A – sin² B = sin (A + B) sin (A – B) 

LHS = sin²(A + B) – sin²(A – B) = sin [(A + B) + (A – B)] [sin (A + B) – (A – B)] 

= sin 2A sin 2B = RHS

(iv) LHS = cos 8θ cos 2θ 

= cos (5θ + 3θ) cos (5θ – 3θ). 

We know cos (A + B) cos (A – B) = cos² A – sin² B 

∴ cos (5θ + 3θ) cos (5θ – 3θ) = cos² 5θ – sin² 3θ = RHS

LHS = cos² A + cos² B – 2 cos A cos B [cos A cos B – sin A sin B] 

= cos² A + cos² B – 2 cos² A cos² B + 2 sin A cos A sin B cos B 

= (cos² A – cos² A cos² B) + (cos² B – cos² A cos² B) + 2 sin A cos A sin B cos B 

= cos² A (1 – cos² B) + cos² B (1 – cos² A) + 2 sin A cos A sin B cos B 

= cos² A sin² B + cos² B sin² A + 2 sin A cos B sin B cos A 

= (sin A cos B + cos A sin B)² 

= sin² (A + B) = RHS

Given cos(α – β) + cos(β – γ) + cos(γ – α) = -3/2

2 cos (α – β) + 2cos (β – γ) + 2cos (γ – α) = -3 

2cos(α – β) + 2cos(β – γ) + 2cos (γ – α) + 3 = 0 

[2 cos α cos β + 2 sin α sin β] + [2 cos β cos γ + 2 sin β sin γ] + [2 cos γ cos α + sin γ sin α] + 3 = 0 

= [2 cos α cos β + 2 cos β cos γ + 2 cos γ cos α] + [2 sin α sin β + 2 sin β sin γ + 2 sin γ sin α] + (sin² α + cos² α) + (sin² β + cos² β) + (sin² γ + cos² γ) = 0 

⇒ (cos² α + cos² β + cos² γ + 2 cos α cos β + 2 cos β cos γ + 2 cos γ cos α) + (sin² α + sin² β) + (sin² γ + 2 sin α sin β + 2 sin β sin γ + 2 sin γ sin α) = 0 

(cos α + cos β + cos γ)² + (sin α + sin β + sin γ)² = 0 

=(cos α + cos β + cos γ) = 0 and sin α + sin β + sin γ = 0 

Hence proved

Correct option is (C) 1-q²/2q 

cosec θ – cot θ = q ……(i) 

cosec² θ – cot² θ = 1 

∴ (cosec θ + cot θ) (cosec θ – cot θ) = 1 

∴ (cosec θ + cot θ)q = 1 

∴ cosec θ + cot θ = 1/q …….(ii) 

Subtracting (i) from (ii), we get

2cot θ = 1/q - q

∴ cot θ = 1-q²/2q 

Given: X + Y = 90°

⇒ X= 90° - Y

Multiplying both side by cos, we get

= cos X = Cos (90° - Y)

⇒ cos X = sin Y [∵ Sin θ = cos (90° - θ)]

(i) Here, A+B = 90°

⇒ A = 90° - B

Multiplying both sides by Sin, we get

Sin A = Sin (90° - B)

⇒ sin A = Cos B [∵ cos θ = sin (90° - θ)]

(ii) Here, A+B = 90°

⇒ B = 90° - A

Multiplying both sides by cos, we get

Cos B = cos (90° - A)

⇒ cos B = sin A [∵ Sin θ = cos (90° - θ)]

(iii) Here, A+B = 90°

⇒ A = 90° - B

Multiplying both sides by sec, we get

Sec A = Sec (90° - B)

⇒ sec A = Cosec B [∵ cosec θ = sec (90° - θ)]

(iv) Here, A+B = 90°

⇒ B = 90° - A

Multiplying both sides by tan, we get

tan B = tan (90° - A)

⇒ tan B = cot A [∵ cot θ = tan (90° - θ)]

(v) Here, A+B = 90°

⇒ B = 90° - A

Multiplying both sides by cosec, we get

Cosec B = cosec (90° - A)

⇒ cosec B = sec A [∵ sec θ = cosec (90° - θ)]

(vi) Here, A+B = 90°

⇒ A = 90° - B

Multiplying both sides by Sin, we get

cotA = cot (90° - B)

⇒ cot A = tan B [∵ tan θ = cot (90° - θ)]

Given: A + B = 90°

⇒ B = 90° - A

Multiplying both side by cos, we get

= cos B = Cos (90° - A)

⇒ cos B = sin A [∵ Sin θ = cos (90° - θ)]

(i) sin 300° = sin(360° – 60°) 

[For 360° – 60°. No change in T-ratio. 300° lies in 4th quadrant ‘sin’ is negative] 

= -sin 60°

= - \(\frac{\sqrt{3}}{2}\)

(ii) cos(-210°) = cos 210° (∵ cos(-θ) = cos θ) 

[∵ 180 + 30°. No change in T-ratio. 210° lies 3rd quadrant ‘cos’ is negative] 

= cos(180° + 30°) 

= -cos 30°

= - \(\frac{\sqrt{3}}{2}\)

(iii) sec 390° = sec(360° + 30°) 

= sec 30° 

= \(\frac{1}{cos 30°}\)

= \(\frac{1}{\frac{\sqrt{3}}{2}}\)

= \(\frac{2}{\sqrt{3}}\)

(iv) tan(-855°) = -tan 855° (∵ tan(-θ) = – tan θ) 

[∵ Multiplies of 360° are dropped out. For 180° – 45°. No change in T-ratio. 180° – 45° lies in 2nd quadrant ‘tan’ is negative] 

= -tan(2 x 360° + 135°) 

= -tan 135° 

= -tan(180° – 45°) 

= -(-tan 45°) 

= -(-1) 

= 1

(v) cosec 1125° = cosec(3 x 360°+ 45°) 

= cosec 45° 

= \(\frac{1}{sin45°}\)

= \(\frac{1}{\frac{1}{\sqrt{2}}}\)

= √2

tan 225° = tan (180° + 45°) 

= tan 45° 

= 1 .

(i) Using transformation formulae 

sin5A . cos3A = \(\frac{1}{2}\)[sin(5A +3A) + sin(5A – 3A)] 

= \(\frac{1}{2}\)(sin8A + sin2A) 

(ii) cos4A – sin2A = \(\frac{1}{2}\)[sin 6A – sin2A] 

(iii) cos \(\frac{5\theta}{2}\). sin\(\frac{\theta}{2}\) = \(\frac{1}{2}\)[sin3θ – sin2θ] 

(iv) 2cos70° . cos10° = 2 x \(\frac{1}{2}\)[cos80° + cos60°] 

= cos80° +\(\frac{1}{2}\)

i. sin θ < 0 sin θ is negative in 3rd and 4th quadrants, tan 0 > 0 

tan θ is positive in 1st and 3rd quadrants. 

∴ θ lies in the 3rd quadrant.

ii. cos θ < 0 cos θ is negative in 2nd and 3rd quadrants, tan 0 > 0 

tan θ is positive in 1st and 3rd quadrants. 

∴ θ lies in the 3rd quadrant.

1899° = 5 x 360° + 99° 

∴ 1899° and 99° are co-terminal angles. 

Since 90° < 99° < 180°, 

99° lies in the 2nd quadrant. 

∴ 1899° lies in the 2nd quadrant. 

∴ cot 1899° is negative.

986° = 2x 360° + 266° 

∴ 986° and 266° are co-terminal angles. 

Since 180° < 266° < 270°, 266° lies in the 3rd quadrant. 

∴ 986° lies in the 3rd quadrant. 

∴ sin 986° is negative.

Since 0° < 4° < 90°, 4° lies in the first quadrant. 

∴ cos4° >0 …(i) 

Since 1^c = 57° nearly, 

180° < 4^c < 270° 

∴ 4^c lies in the third quadrant. 

∴ cos 4^c < 0 ………(ii) 

From (i) and (ii), 

cos 4° is greater.

(a) 1

tan (– 1575°) 

= – tan (1575°)                (∵ tan (–θ) = – tanθ) 

= – tan (4 × 360° + 135°) 

= – tan (135°)             (∵ tan (n.360° + θ) = tan θ) 

= – tan (90° + 45°)          (∵ tan (90° + θ) = – cot θ) 

= – (– cot 45°) = cot 45° = 1.

(a) -1

cos 1° + cos 2° + cos 3° + ........ + cos 180° 

= (cos 1° + cos 179°) + (cos 2° + cos 178°) + ........ + (cos 89° + cos 91°) + cos 90° + cos 180º 

= [cos 1° + cos (180° – 1°)] + [cos 2° + cos (180° – 2°)] + ....... + [cos 89° + cos (180° – 91°)] + cos 90° + cos 180° 

= (cos 1° – cos 1°) + (cos 2° – cos 2°) + ....... + (cos 89° – cos + 89°) + 0 + (– 1) = – 1. 

(∵ cos (180° – θ) = – cos θ)

(d) 0

In a cyclic quadrilateral, the sum of the opposite angles is 180°. 

⇒ A + C = 180° and B + D = 180° 

∴ cos A + cos B + cos C + cos D 

= cos A + cos B + cos (180° – C) + cos (180° – B) 

= cos A + cos B – cos A – cos B 

= 0.

= 2sin (\(\frac{6x+4x}{2}\)) cos (\(\frac{6x-4x}{2}\)) – 2 sin x cos x

= 2 sin 5x cos x — 2 sin x cos x

= 2 cos x (sin 5x — sin x)

= 2 cos[2 cos(\(\frac{5x+x}{2}\)) sin (\(\frac{5x-x}{2}\))]

= 2 cos x (2 cos 3x sin 2x) 

= 4 cos x sin 2x cos 3x 

= R.H.S.

(a) \(\frac18\)

\(\bigg(1+cos\fracπ8\bigg)\)\(\bigg(1+cos\frac{3π}8\bigg)\)\(\bigg(1+cos\frac{5π}8\bigg)\)\(\bigg(1+cos\frac{7π}8\bigg)\)

= \(\bigg(1+cos\fracπ8\bigg)\)\(\bigg(1+cos\frac{3π}8\bigg)\)\(\bigg(1+cos\bigg(π-cos\frac{3π}8\bigg)\bigg)\)\(\bigg(1+cos\bigg(π-\frac{π}8\bigg)\bigg)\)

=  \(\bigg(1+cos\fracπ8\bigg)\)\(\bigg(1+cos\frac{3π}8\bigg)\)\(\bigg(1-cos\frac{3π}8\bigg)\)\(\bigg(1-cos\frac{π}8\bigg)\)        (∵cos (π – θ) = – cos θ)

= \(\bigg(1-cos^2\fracπ8\bigg)\)\(\bigg(1-cos^2\frac{3π}8\bigg)\) = sin² \(\fracπ8\)  sin² \(\frac{3π}8\)

=  sin² \(\fracπ8\)  cos² \(\frac{π}8\) \(\bigg(∵ sin\frac{3π}{8}= sin\bigg(\fracπ2-\fracπ8\bigg)=cos\fracπ8\bigg)\)

= \(\frac14\bigg(4\,sin^2\fracπ8cos^2\fracπ8\bigg)\) = \(\frac14\bigg(2\,sin\fracπ8\,cos\fracπ8\bigg)^2\)

= \(\frac14\big(sin\fracπ4\big)^2\,\frac14\big(\frac1{\sqrt2}\big)^2=\frac18.\)             (Using, 2 sin θ cos θ = sin 2θ)

cot 225° = \(\frac{1}{ tan 225°}=\frac{1}{tan(180°+45°)}\)

= \(\frac{1}{(\frac{tan180°+tan 45°}{1-tan180° tan 45°})}\)

= \(\frac{1}{(\frac{0+1}{1-0(1)})}\)

= \(\frac{1}{(\frac{1}{1})}\)

=1

cot (-1110°) =-cot (1110°) 

= -cot (1080°+ 30°) 

= – cot (3 x 360° + 30° ) 

= – cot 30°

= √3

cos 600° = cos (360° + 240°) 

= cos 240° 

= cos (180° + 60°) 

= – cos 60°

= - 1/2

cos 315° = cos (270° + 45°)

sin 45° = 1/√2

(d) \(\frac{-1}{2}\)

cos(-480°) = cos 480° [∵ cos(-θ) = cos θ] 

= cos(360° + 120°) 

= cos 120° 

= cos(180° – 60°)

= -cos 60° 

= \(\frac{-1}{2}\)

(d) √3+1

2x² cos 60° – 4 cot² 45° – 2 tan 60° = 0 

⇒ 2x² x \(\frac12\) - 4 x (1)² - 2 x √3 = 0 

⇒ x² - 4 - 2√3 = 0 ⇒ x² = 4 + 2√3

⇒ x² = 3 + 1 + 2√3 ⇒ x² = (√3)² + (1)² + 2√3

⇒ x² = (√3 + 1)² ⇒ x = √3+1.

(b) G.P

cot \(\fracπ3\) = cot 60° = \(\frac{1}{\sqrt3}\), + cot \(\fracπ4\) = cot 45° = 1, cot \(\fracπ6\) = cot 30° = √3

We can see that

\(\bigg(cot \fracπ3\bigg).\bigg(cot\bigg(\frac{π}{6}\bigg)\bigg)=\frac{1}{\sqrt3}.\sqrt3=1=\bigg(cot\frac{π}{4}\bigg)^2\)

∴ cot \(\fracπ3\), cot \(\fracπ4\), cot \(\fracπ6\) are in G.P

2 sin 2θ = √3

=> sin(2θ) = √3/2 

=> sin(2θ) = sin(60°) 

=> 2θ = 60° 

=> θ = 60°/2 

=> θ = 30°

Correct answer is (B) 0

Explanation : 

cos 1° cos 2° cos 3° … cos 179° 

= cos 1° cos 2° cos 3° … cos 90°… cos 179° 

= 0 …[∵ cos 90° = 0]

\(\frac{cos^3 20° - cos^3 70°}{sin^3 70° - sin^3 20°}\) = \(\frac{cos^3 20° - cos^3 (90°-20°)}{sin^3 (90°-20°) - sin^3 20°}\)

\(\frac{cos^3 20° - sin^3 20°}{cos^3 20° - sin^3 60°}\) = 1

Solution :

L.H.S. = (sin⁴θ – cos⁴θ +1) cosec²θ

= [(sin²θ – cos²θ) (sin²θ + cos²θ) + 1] cosec²θ

= (sin²θ – cos²θ + 1) cosec²θ

[Because sin ²θ + cos²θ =1]

= 2sin²θ cosec²θ [Because 1– cos ²θ = sin²θ ]

= 2 = RHS

Answer : (c)  nπ ± \(\frac{\pi}{8}\)

 sec² x = \(\sqrt{2}\)  (1 - tan² α) 

⇒ tan² α  + 1 = \(\sqrt{2}\) (1 – tan² α )  

⇒ tan² α (1 + \(\sqrt{2}\) ) = \(\sqrt{2}\) – 1 

⇒ tan² α  = \(\frac{\sqrt{2} -1}{\sqrt{2} +1}\) = \(\frac{(\sqrt{2} -1)^2}{(\sqrt{2}+1)(\sqrt{2}-1)}\)  = \((\sqrt{2}-1)^2\) = \({tan}^2\,\frac{\pi}{8}\) 

∴ tan α = tan \(\big(± \frac{\pi}{8}\big)\) 

α = nπ ± \(\frac{\pi}{8}\)

Answer

(i) (B) is correct.

9 sec²A - 9 tan²A

= 9 (sec²A - tan²A)
= 9×1 = 9             (∵ sec2 A - tan2 A = 1)

(ii) (C) is correct

(1 + tan θ + sec θ) (1 + cot θ - cosec θ)   

= (1 + sin θ/cos θ + 1/cos θ) (1 + cos θ/sin θ - 1/sin θ)

= (cos θ+sin θ+1)/cos θ × (sin θ+cos θ-1)/sin θ

= (cos θ+sin θ)²-1²/(cos θ sin θ)

= (cos²θ + sin²θ + 2cos θ sin θ -1)/(cos θ sin θ)

= (1+ 2cos θ sin θ -1)/(cos θ sin θ)

= (2cos θ sin θ)/(cos θ sin θ) = 2

(iii) (D) is correct.

(secA + tanA) (1 - sinA)

= (1/cos A + sin A/cos A) (1 - sinA)

= (1+sin A/cos A) (1 - sinA)

= (1 - sin²A)/cos A

= cos²A/cos A = cos A

(iv) (D) is correct.

1+tan²A/1+cot²A 

= (1+1/cot²A)/1+cot²A

= (cot²A+1/cot²A)×(1/1+cot²A)

= 1/cot²A = tan²A

2sin 3x = √3

⇒ sin 3x = \(\frac{\sqrt{3}}{2}\)

⇒ sin 3x = sin 60°

⇒ 3x = 60°

⇒ x = 20°

(sec θ + cos θ) (sec θ – cos θ) = sec² θ – cos² θ 

= (1 + tan² θ) – (1 – sin² θ) 

= tan² θ + sin² θ = RHS

Taking A + B = X and C = Y 

We get cos (X + Y) = cos X cos Y – sin X sin Y 

(i.e) cos (A + B + C) = cos (A + B) cos C – sin (A + B) sin C 

= (cos A cos B – sin A sin B) cos C – [sin A cos B + cos A sin B] sin C 

cos (A + B + C) = cos A cos B cos C – sin A sin B cos C – sin A cos B sin C – cos A sin B sin C 

If (A + B + C) = π/2 then cos (A + B + C) = 0 

⇒ cos A cos B cos C – sin A sin B cos C – sin A cos B sin C – cos A sin B sin C = 0 

⇒ cos A cos B cos C = sin A sin B cos C + sin B sin C cos A + sin C sin A cos B

cos 204° = cos (180°+ 24°) = – cos 24° 

cos 125° = cos (180° – 55°) = – cos 55° 

LHS = cos 24° + cos 55° + (- cos 55°) + (- cos 24°) + cos 300° 

= cos 24° + cos 55° – cos 55° – cos 24° + cos 300°

= cos 300° = cos(360° - 60°) = cos 60° = 1/2 = RHS

LHS = sin (270° – θ) sin (90° – θ) – cos (270° – θ) cos (90° + θ) + 1 

Now, sin (270° – θ) = sin {180°+ (90°- θ)} 

= – cos (90° – θ) = – sin θ 

LHS = – cos θ . cos θ – (- sin θ) (- sin θ) + 1

= – cos² θ – sin² θ + 1 

= – (cos² θ + sin² θ) + 1 = -1 + 1 = 0 = RHS

sec(α + π/2) = - cosec α

LHS = [(1 + cot α) + cosec α][(1 + cot α) – cosec α] 

= (1 + cot α)² – cosec² α 

= 1 + cot² α + 2 cot α – cosec² α 

[∵ 1 + cot² α = cosec² α]

= cosec² α + 2 cot α – cosec² α 

= 2 cot α = RHS

Let θ = 18° 

∴ 5θ = 90°

∴ 2θ + 3θ = 90° 

∴ 2θ = 90° – 3θ 

∴ sin 2θ = sin (90° – 3θ) 

∴ sin 2θ = cos 3θ 

∴ 2 sin θ cos θ = 4 cos³ θ – 3 cos θ 

∴ 2 sin θ = 4 cos² θ – 3 …..[∵ cos θ ≠ 0] 

∴ 2 sin θ = 4 (1 – sin² θ) – 3 

∴ 2 sin θ = 1 – 4 sin² θ 

∴ 4 sin² θ + 2 sin θ – 1 = 0

∴ sin θ = \(\frac{-2±\sqrt{4+16}}{8}\)

=\(\frac{-2±2\sqrt5}{8}\)

=\(\frac{-1±\sqrt5}{4}\)

Since, sin 18° > 0

∴ sin 18°=\(\frac{\sqrt5-1}{4}\)

(sin x – cos x) sin⁶ x + cos⁶ x

= (sin² x)³ + (cos² x)³ 

= (sin² x + cos² x)³ – 3 sin² x cos² x (sin² x + cos² x) ….. [∵ a³ + b³ = (a + b)³ – 3ab(a + b)] 

= 1³ – 3 sin² x cos² x (1) 

= 1 – 3 sin² x cos² x 

L.H.S. = 3(sin x – cos x)⁴ + 6(sin x + cos x)² + 4(sin⁶ x + cos⁶ x) 

= 3(1 – 4 sin x cos x + 4 sin² x cos² x) + 6(1 + 2 sin x cos x) + 4(1 – 3 sin² x cos² x) 

= 3 – 12 sin x cos x + 12 sin² x cos² x + 6 + 12 sin x cos x + 4 – 12 sin² x cos² x 

= 13 

= R.H.S.

sin 495° = sin (360° + 135°) 

= sin (135°) 

= sin (90° + 45°) 

= cos 45°

= 1/√2

sin 690° = sin (720° -30°) 

= sin (2 x 360° – 30°) 

= – sin 30°

= -1/2

tan (- 690°) = – tan 690° 

= – tan (720° – 30°) 

= – tan (2 x 360° – 30°) 

= – (- tan 30°) 

= tan 30°

= 1/√3

tan 105° = tan (60° +45°)

= \(\frac{tan60°+tan45°}{1-tan60° tan 45°}\)

=\(\frac{\sqrt3+1}{1-(\sqrt3)(1)}\)

=\(\frac{\sqrt3+1}{1-\sqrt3}\)

cos 75° = cos (45° + 30°) 

= cos 45° cos 30° – sin 45° sin 30°

=(1/√2) (√3/2) - (1/√2) (1/2) 

= \(\frac{\sqrt3-1}{2\sqrt2}\)

Given, 

cos 75° + cot 75° 

Since, cos (90 – θ) = sin θ and cot (90 – θ) = tan θ 

cos 75° + cot 75° 

= cos (90 – 15)° + cot (90 – 15)° 

= sin 15° + tan 15° 

Hence, cos 75° + cot 75° can be expressed as sin 15° + tan 15°.

We know that, 

sin (90 – θ) = cos θ and cos (90 – θ) = sin θ 

So, the given can be expressed as 

sec 50° sin (90 – 50)° + cos (90 – 50)° cosec 50° 

= sec 50° cos 50° + sin 50° cosec 50° 

= 1 + 1 [∵ sin θ x cosec θ = 1 and cos θ x sec θ = 1] 

= 2

Given

Sin A = \(\frac{1}{2}\)

Sin A = Sin 30°

Now,

A = 30° ----(1)

Cos B =  \(\frac{1}{2}\)

Cos B = Cos 60°

B = 60° ----(2)

Now,

Value of (A + B)

= 30° + 60° [ from (1)&(2) ]

= 90°

Therefore,.

A + B = 90°