Trigonometry
- If tan4θ + tan2θ = 1 then the value of cos4θ + cos2θ is
-
View Hint View Answer Discuss in Forum
tan4θ + tan2θ = 1
⇒ tan2θ (tan2θ + 1) = 1
⇒ tan2θ . sec2θ = 1⇒ tan2θ = 1 = cos2θ. sec2θ
∴ cos4θ + cos2θ = tan4θ + tan2θ = 1Correct Option: C
tan4θ + tan2θ = 1
⇒ tan2θ (tan2θ + 1) = 1
⇒ tan2θ . sec2θ = 1⇒ tan2θ = 1 = cos2θ. sec2θ
∴ cos4θ + cos2θ = tan4θ + tan2θ = 1
- (1 + sec 20° + cot 70°)(1 – cosec 20° + tan70°) is equal to
-
View Hint View Answer Discuss in Forum
(1 + sec 20° + cot 70°) (1 – cosec 20° + tan 70°)
= (1 + sec 20° + tan 20°) (1 – cosec 20° + cot 20°)
[∴ tan (90° – θ) = cotθ; cot (90° – θ) = tanθ]= 1 + 1 + sin20° 1 − 1 + cos20° cos20° cos20° sin20° sin20° = cos20° + 1 + sin20° sin20° – 1 + cos20° cos20° sin20° = (sin20° + cos20° + 1)( sin20° + cos20° + 1) sin20°.cos20° = (sin20° + cos20°)2 − 1 sin20°.cos20° = sin220° + cos220° + 2sin20°.cos20°− 1 sin20°.cos20° = 1 + 2sin20°.cos20°− 1 = 2 sin20°.cos20° Correct Option: C
(1 + sec 20° + cot 70°) (1 – cosec 20° + tan 70°)
= (1 + sec 20° + tan 20°) (1 – cosec 20° + cot 20°)
[∴ tan (90° – θ) = cotθ; cot (90° – θ) = tanθ]= 1 + 1 + sin20° 1 − 1 + cos20° cos20° cos20° sin20° sin20° = cos20° + 1 + sin20° sin20° – 1 + cos20° cos20° sin20° = (sin20° + cos20° + 1)( sin20° + cos20° + 1) sin20°.cos20° = (sin20° + cos20°)2 − 1 sin20°.cos20° = sin220° + cos220° + 2sin20°.cos20°− 1 sin20°.cos20° = 1 + 2sin20°.cos20°− 1 = 2 sin20°.cos20°
- If a2 sec2x – b2 tan2x = c2 then the
value of (sec2x + tan2x) is equal to (assume b2 ≠ a2)
-
View Hint View Answer Discuss in Forum
a2 sec2x – b2 tan2x = c2
⇒ a2 (1 + tan2x) – b2 tan2x = c2
⇒ a2 + a2tan2x – b2 tan2x = c2
⇒ a2tan2x – b2 tan2x = c2 – a2
⇒ tan2x(a2 – b2) = c2 – a2
⇒ tan2x = c2 – a2 a2 – b2
∴ sec2x + tan2x
= 1 + tan2x + tan2x
= 1 + 2 tan2x= 1 + 2(c2 – a2) a2 – b2 = a2 – b2 + 2c2 – 2a2 a2 – b2 = – b2 + 2c2 – a2 a2 – b2 = b2 + a2 – 2c2 b2 – a2 Correct Option: B
a2 sec2x – b2 tan2x = c2
⇒ a2 (1 + tan2x) – b2 tan2x = c2
⇒ a2 + a2tan2x – b2 tan2x = c2
⇒ a2tan2x – b2 tan2x = c2 – a2
⇒ tan2x(a2 – b2) = c2 – a2
⇒ tan2x = c2 – a2 a2 – b2
∴ sec2x + tan2x
= 1 + tan2x + tan2x
= 1 + 2 tan2x= 1 + 2(c2 – a2) a2 – b2 = a2 – b2 + 2c2 – 2a2 a2 – b2 = – b2 + 2c2 – a2 a2 – b2 = b2 + a2 – 2c2 b2 – a2
- x, y be two acute angles, x + y < 90° and sin(2x – 20°) = cos (2y + 20°), the value of tan (x + y) is
-
View Hint View Answer Discuss in Forum
sin (2x – 20°) = cos (2y + 20°)
⇒ sin (2x – 20°)
= sin {90° – (2y + 20°)}
⇒ 2x – 20° = 90° – 2y – 20°
⇒ 2x + 2y = 90°
⇒ 2 (x + y) = 90° ⇒ x + y = 45°
∴ tan (x + y) = tan 45° = 1Correct Option: C
sin (2x – 20°) = cos (2y + 20°)
⇒ sin (2x – 20°)
= sin {90° – (2y + 20°)}
⇒ 2x – 20° = 90° – 2y – 20°
⇒ 2x + 2y = 90°
⇒ 2 (x + y) = 90° ⇒ x + y = 45°
∴ tan (x + y) = tan 45° = 1
- If r sinθ = √3 r cosθ = 1, rhen values of r and θ are : (0° ≤ θ ≤ 90°)
-
View Hint View Answer Discuss in Forum
r sinθ = √3
r cosθ = 1
On squaring and adding,
r2sin2θ + r2cos2θ = 3 + 1
⇒ r2 (sin2θ + cos2θ) = 4Again, r sinθ = √3 r cosθ
⇒ tanθ = √3 = tan 60°
⇒ θ = 60°Correct Option: D
r sinθ = √3
r cosθ = 1
On squaring and adding,
r2sin2θ + r2cos2θ = 3 + 1
⇒ r2 (sin2θ + cos2θ) = 4Again, r sinθ = √3 r cosθ
⇒ tanθ = √3 = tan 60°
⇒ θ = 60°