Trigonometry

11 )

If [(sinx) / a] = [(cosx) / b] = [(tanx) / c] = K then bc + (1 / ck) + [ak / (1 + bk)] = ________

k[a + (1/a)]

(1/k)[a + (1/a)]

(1 / k^2)

(a/k)

View Solution

Answer :

[(sinx) / a] = [(cosx) / b] = [(tanx) / c] = k

given:

bc + (1 / ck) + [ak / (1 + bk)]

= [{(cosx) / k} × {(tanx) / k}] + [1 / {k × [(tanx) / k]}]

+ [{[(sinx) / k] × k} / {1 + [(cosx) / k] × k}]

= (1 / k²)sinx + [(cosx) / (sinx)] + [(sinx) / (1 + cosx)]

= [(sinx) / k²] + [(cosx + cos²x + sin²x) / {sinx(1 + cosx)}]

= [(sinx) / k²] + [1 / (sinx)]

= (ka / k²) + (1 / ak)

= (a/k) + (1 / ak)

= (1/k)[a + (1/a)].

12 )

If cosx = 1 – 2sin²32°, α, β are the value of x between 0° and 360° with α < β then α = ________

180° – β

200° – β

(β/4) – 10°

(β/5) – 4°

View Solution

Answer :

cosx = 1 – 2sin²32

∴ cosx = cos(2 × 32)

∴ cosx = cos64

∴ x = 64° or x = 296°

as α < β

∴ α = 64° and β = 296°

∴ α = [(296) / 4] – 10°

∴ α = (β/4) – 10°.

13 )

The minimum value of 125tan²θ + 5cot²θ is

105

View Solution

Answer :

given: 125tan²θ + 5cot²θ

if the expression is of type

asin²θ + bcosec²θ

acos²θ + bsec²θ

atan²θ + bcot²θ

then minimum value is 2√(ab).

here a = 125 and b = 5.

∴ minimum value is given by 2√(125 × 5) = 50.

14 )

If A = cos⁴θ + sin²θ, ∀θ ∈ R then A lies in the interval

[1, 2]

[(3/4), 1]

[(13 / 16), 1]

[(3/4), (13 / 16)]

View Solution

Answer :

A = cos⁴θ + sin²θ

= sin²θ + (1 – sin²θ)²

= 1 + sin²θ + sin⁴θ – 2sin²θ

= 1 + sin²θ(sin²θ – 1)

= 1 + sin²θ(– cos²θ)

= 1 – sin²θ ∙ cos²θ

∴ A = (1 – sin²θ ∙ cos²θ) ≤ 1 (1)

also A = 1 – [(sin²2θ) / 4] ≥ [1 – (1/4)]

i.e. A = 1 – [(sin²2θ) / 4] ≥ (3/4) (2)

∴ from (1) & (2)

(3/4) ≤ A ≤ 1.

15 )

	sin²x	cos²x	1
A =	cos²x	sin²x	1
	– 10	12	2

then A = ________

10sin^2 x

12cos^2 x – 10sin^2 x

12cos^2 x

View Solution

Answer :

	sin²x	cos²x	1
A =	cos²x	sin²x	1
	– 10	12	2

∴ A = sin²x(2sin²x – 12) – cos²x(2cos²x + 10) + 1(12cos²x + 10sin²x)

∴ A = 2sin⁴x – 12sin²x – 2cos⁴x – 10cos²x + 12cos²x + 10sin²x

= 2(sin⁴x – cos⁴x) – 2sin²x + 2cos²x

= 2[(sin⁴x – cos⁴x) – (sin²x – cos²x)]

= 2(sin²x – cos²x) [(sin²x + cos²x) – 1]

= 0.

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