Trigonometry
- The value of (cos 53° – sin 37°) is
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cos53° – sin37°
= cos (90° – 37°) – sin37°
= sin37° – sin37° = 0Correct Option: A
cos53° – sin37°
= cos (90° – 37°) – sin37°
= sin37° – sin37° = 0
- If cosecθ + sinθ = 5/2, then the value of (cosecθ – sinθ) is :
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cosecθ + sinθ = 5 2 ⇒ 1 + sinθ = 5 2 sinθ ⇒ 1 + sinθ = 5 2 sinθ
⇒ 2 sin²θ + 2 = 5sinθ
⇒ 2 sin2θ – 5 sinθ + 2 = 0
⇒ 2 sin2θ – 4 sinθ – sinθ + 2 = 0
⇒ 2 sinθ (sinθ – 2) – 1 (sinθ – 2) = 0
⇒ (2 sinθ – 1) (sinθ – 2) = 0
⇒ 2 sinθ – 1 = 0
⇒ 2 sinθ = 1⇒ sinθ = 1 because sinθ ≠ 2 2
⇒ cosecθ = 2∴ cosecθ – sinθ = 2 – 1 = 3 2 2 Correct Option: B
cosecθ + sinθ = 5 2 ⇒ 1 + sinθ = 5 2 sinθ ⇒ 1 + sinθ = 5 2 sinθ
⇒ 2 sin²θ + 2 = 5sinθ
⇒ 2 sin2θ – 5 sinθ + 2 = 0
⇒ 2 sin2θ – 4 sinθ – sinθ + 2 = 0
⇒ 2 sinθ (sinθ – 2) – 1 (sinθ – 2) = 0
⇒ (2 sinθ – 1) (sinθ – 2) = 0
⇒ 2 sinθ – 1 = 0
⇒ 2 sinθ = 1⇒ sinθ = 1 because sinθ ≠ 2 2
⇒ cosecθ = 2∴ cosecθ – sinθ = 2 – 1 = 3 2 2
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The value of 2tan 53° - cot 80° is : cot37 ° tan 10 °
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2tan 53° - cot 80° cot 37° tan 10° = 2tan (90° - 37°) - cot (90° - 10°) cot 37° tan 10° = 2cot 37° - tan 10° cot 37° tan 10°
= 2 – 1 = 1Correct Option: C
2tan 53° - cot 80° cot 37° tan 10° = 2tan (90° - 37°) - cot (90° - 10°) cot 37° tan 10° = 2cot 37° - tan 10° cot 37° tan 10°
= 2 – 1 = 1
- The value of cot 10°. cot 20°. cot 60°. cot 70°. cot 80° is :
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Expression = cot10°.cot20°.cot60°.cot70°.cot80° = (cot10°.cot80°) (cot20°. cot70°).cot60°
= {cot10°.cot (90°–10°)} {cot20°.cot(90° – 20°)}. 1 = (cot10°.tan10°)(cot20°.tan20°). 1 √3 √3 = 1.1 1 = 1 √3 √3
[∵ cot (90° – θ) = tanθ; tanθ.cotθ = 1]Correct Option: D
Expression = cot10°.cot20°.cot60°.cot70°.cot80° = (cot10°.cot80°) (cot20°. cot70°).cot60°
= {cot10°.cot (90°–10°)} {cot20°.cot(90° – 20°)}. 1 = (cot10°.tan10°)(cot20°.tan20°). 1 √3 √3 = 1.1 1 = 1 √3 √3
[∵ cot (90° – θ) = tanθ; tanθ.cotθ = 1]
- If 7sin²θ + 3cos²θ = 4, and 0° < θ < 90°, then the value of tanθ is :
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7 sin²θ + 3 cos²θ = 4
On dividing by cos²θ,7 sin²θ + 3 cos²θ = 4 cos²θ cos²θ cos²θ
⇒ 7tan²θ + 3 = 4 sec²θ = 4 (1 + tan²θ)
⇒ 7tan²θ + 3 = 4 + 4 tan2θ
⇒ 7tan²θ – 4 tan2θ = 4 – 3
⇒ 3tan²θ = 1⇒ tan²θ = 1 3 ⇒ tan²θ = 1 √3 Correct Option: B
7 sin²θ + 3 cos²θ = 4
On dividing by cos²θ,7 sin²θ + 3 cos²θ = 4 cos²θ cos²θ cos²θ
⇒ 7tan²θ + 3 = 4 sec²θ = 4 (1 + tan²θ)
⇒ 7tan²θ + 3 = 4 + 4 tan2θ
⇒ 7tan²θ – 4 tan2θ = 4 – 3
⇒ 3tan²θ = 1⇒ tan²θ = 1 3 ⇒ tan²θ = 1 √3