Trigonometry
-  If tan4θ + tan2θ = 1 then the value of cos4θ + cos2θ is
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                        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: Ctan4θ + 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
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                        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)
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                        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: Ba2 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
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                        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: Csin (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°)
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                        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: Dr 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°
 
	