Trigonometry
- The simple value of tan 1°. tan 2°. tan 3°....... tan 89° is
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tan 1° . tan 2° . tan 3° ... tan 89°
= (tan 1° . tan 89°) (tan 2°. tan88°) ... tan 45°
= (tan 1°. tan(90° – 1°) ) (tan 2° tan (90° – 2°) .. tan 45°
= (tan 1° . cot 1°) (tan 2°. cot 2°) .. tan 45°
= 1.1..1. = 1
[tan (90° – θ) = cot θ]Correct Option: C
tan 1° . tan 2° . tan 3° ... tan 89°
= (tan 1° . tan 89°) (tan 2°. tan88°) ... tan 45°
= (tan 1°. tan(90° – 1°) ) (tan 2° tan (90° – 2°) .. tan 45°
= (tan 1° . cot 1°) (tan 2°. cot 2°) .. tan 45°
= 1.1..1. = 1
[tan (90° – θ) = cot θ]
- If x sin² 60° – (3 / 2) sec 60° tan² 30° + (4 / 5) sin² 45° tan² 60° = 0 then x is
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Correct Option: C
x sin²60° – 3 sec 60° . tan²30° 2 + 4 sin²45° . tan² 60° = 0 5 
⇒ 3x - 3 × 2 × 1 + 4 × 1 × 3 = 0 4 2 3 5 2 ⇒ 3x - 1 + 6 = 0 4 5 ⇒ 3x = 1 - 6 = 5 - 6 = - 1 4 5 5 5 ⇒ x = - 1 × 4 = - 4 5 3 15
- If 7 sinα = 24 cos α; 0 < α < (π / 2) , then the value of 14 tan α – 75 cos α – 7 sec α is equal to
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7 sin α = 24 cos α
⇒ sin α = 24 ⇒ tanα = 24 cosα 7 7 
∴ cosα = 1 = 7 sec α 7
∴ 14 tanα – 75 cosα – 7 secα= 14 × 24 - 75 × 7 - 7 × 25 7 25 7
= 48 – 21 – 25 = 2Correct Option: D
7 sin α = 24 cos α
⇒ sin α = 24 ⇒ tanα = 24 cosα 7 7 ∴ cosα = 1 = 7 sec α 7
∴ 14 tanα – 75 cosα – 7 secα= 14 × 24 - 75 × 7 - 7 × 25 7 25 7
= 48 – 21 – 25 = 2
- The value of x which satisfies the equation 2 cosec² 30° + x sin² 60° – (3 / 4) tan² 30° = 10 is
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2 cosec²30° + x sin²60° - 3 tan²30° = 10 4 
⇒ 8 + 3x - 3 × 1 = 10 4 4 3 ⇒ 3x = 9 ⇒ 3x = 9 4 4 ⇒ x = 9 = 3 3
Correct Option: B
2 cosec²30° + x sin²60° - 3 tan²30° = 10 4 ⇒ 8 + 3x - 3 × 1 = 10 4 4 3 ⇒ 3x = 9 ⇒ 3x = 9 4 4 ⇒ x = 9 = 3 3
- If 2 sin θ + cos θ = (7 / 3) then the value of (tan² θ – sec² θ) is
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tan²θ – sec²θ = – (sec²θ – tan²θ) = –1.
Correct Option: B
tan²θ – sec²θ = – (sec²θ – tan²θ) = –1.
