Question

Solution

Correct option is

t4 > t3 > t1 > t2

θ Ïµ (0,π/4)

0 < tan θ < 1 and cos θ > 1 ⇒ log cot θ > 0

Now t1 = (tan θ)tanθ ⇒ log t1 = tan θ log (tan θ  Similarly log t2 = – cot θ log (cot θ

log t3 = tan θ log (cot θ), log t4 = cot θ log (cot θ)
As cot θ > tan θ, we have

log t4 > log t3 > log t1 > log t2

⇒ t4 > t3 > t1 > t2

SIMILAR QUESTIONS

Q1

If sin θ and cos θ are the roots of the equation ax2 – bx + c = 0, then aband c satisfy the relation

Q2

If sin x + sin2 x = 1, then the value of cos12 x + 3 cos10 x + 3 cos8 x + cos6 x – 1 is equal to

Q3

The value of Q4

If θ lies in the first quadrant and cos θ = 8/17, then the value of cos (300 +θ) + cos (450 – θ) + cos (1200 – θ) is

Q5

If A lies in the second quadrant and 3 tan A + 4 = 0, the value of 2 cot A – 5 cos A + sin A is equal to

Q6

The value of the determinant Is zero if

Q7 is equal to

Q8

An angle α is divided into two parts so that the ratio of the tangents of these parts is λ. If the difference between these parts is x then sinx/sinα is equal to

Q9 or equal to

Q10

If x = sin αy = sin βz = sin (α + β) then cos (α + β) =