## Question

### Solution

Correct option is

0

From sin x + sin2 x = 1, we get sin x = cos2 x. Now, the given expression is equal to

cos6 x (cos6 x + 3 cos4 x + 3 cos2 x + 1) – 1

= cos6 x (cos2 x + 1)3 – 1

= sin3 x (sin x + 1)3 – 1

= (sin2 x + sin x)3 – 1 = 1 – 1 = 0.

#### 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

The value of

Q3

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

Q4

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

Q5

The value of the determinant

Is zero if

Q6

is equal to

Q7

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

Q8

or equal to

Q9

Given θ Ïµ (0,π/4) and t1 = (tan θ)tanθ t2 = (tan θ)cotθt3 = (cot θ)tanθand t4 = (cot θ)cotθ then

Q10

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