A man on the top of a vertical tower observes a car moving at a uniform speed towards the tower on a horizontal road. If it takes 18 min. for the angle of depression of the car to change from $30^{\circ}$ to $45^{\circ}$ ; then after this, the time taken (in min.) by the car to reach the foot of the tower, is :
Let length of tower = h
$\Rightarrow AC' = AB = h$
and $AC = AB cot 30^{\circ}= \sqrt{3} h$$\Rightarrow CC' = (\sqrt{3} -1) h $
Time taken by car form C to C' = 18 min
$\Rightarrow$ time take by car to reach the foot of the tower = $\frac{18}{\sqrt{3}-1}$ min.
= 9 ( $\sqrt{3}$ + 1) min
Cosecant and Secant are even functions, all the others are odd.
sin (-A) = – sinA,
cos (-A) = cos A,
cosec (-A) = -cosec A,
cot (-A) = -cot A,
tan (-A) = – tan A,
sec (-A) = sec A.
Pythagorean Identities
sin2θ + cos2θ = 1
1 + tan2θ = sec2θ
1 + cot2θ = cosec2θ
Periodic Functions
T-Ratios of (2π + x) sin (2π + x) = sin x, cos (2π + x) = cos x, tan (2π + x) = tan x, cosec (2π + x) = cosec x, sec (2π + x) = sec x, cot (2π+x)=cotx.
T-Ratios of (π -x) sin (π–x) = sin x, cos (π–x) = - cos x, tan (π–x) = - tan x, cosec (π–x) = cosec x, sec (π–x) = - sec x, cot (π–x) = - cot x.
T-Ratios of (π+ x) sin (π+x) = - sin x, cos (π+x) = - cos x, tan (π+x) = tan x, cosec (π+x) = - cosec x, sec (π+x) = - sec x, cot (π+x) = cot x.
T-Ratios of (2π – x) sin (2π–x) = - sin x, cos (2n–x) = cos x, tan (2π–x) = - tan x, cosec (2π–x) = - cosec x, sec (2π–x) = sec x, cot (2π-x) = - cot x
Sum and Difference Identities
T-Ratios of (x + y) sin (x+y) = sinx.cosy + cosx.sin y cos (x+y) = cosx.cosy – sinx.siny
T-Ratios of (x – y) sin (x–y) = sinx.cosy – cos.x.sin y cos (x-y) = cosx.cosy + sinx.siny
Product of T-ratios
2sinx cosy = sin(x+y) + sin(x–y)
2cosx siny = sin(x+y) – sin(x–y)
2 cosx cosy = cos(x+y) + cos(x–y)
2sinx.siny = cos(x–y) – cos(x+y)
T-Ratios of (2x) sin2x = 2sin x cos x cos 2x = cos2x – sin2x
= 2cos2x – 1
= 1 – 2sin2x
T-Ratios of (3x) sin 3x = 3sinx – 4sin3x cos 3x = 4cos3x – 3cosx
