2008/010) If P = arctan 1 + 1/2.arctan 2 + 1/3.arctan 3 and Q = arctan 1 + 2 arctan 1/2 + 3 arctan 1/3, then (P - Q)/5 = :

we know arctan 1/x = pi/2 - arc tan x

so Q = arctan 1 + 2(pi/2 - arc tan 2) + 3(pi/2 - arc tan 3)
= 1 + 5pi/2 - 2 arc tan 2 - 3 arc tan 3

so P-Q = 1/2 arc tan 2 + 1/3 arctan 3 -(5pi/2 - 2 arc tan 2 - 3 arc tan 3)
= 5/2 arctan 2 + 10/3 arc tan 3 - 5pi/2

devide by 5 to get
(P-Q)/5 = 1/2 arc tan 2 + 2/3 arc tan 3 - pi/2

= 1/6( 3 arc tan 2 + 4 arc tan 3) - pi/2

3 arctan 2+ 4 arctan 3
= 3 (arc tan 2 + arc tan 3 ) + arctan 3

we know arc tan 2 + arc tan 3 = arctan ( 2+3)/(1-2*3) = arc tan (-1) = - pi/4


so (P-Q)/5 = 1/6(arctan 3 - 3 pi/4)
= 1/6 arctan 3 - pi/8

2008/009) Prove; 1+ cos 56 + cos 58 - cos 66 = 4*cos 28. cos 29. sin 33 ?

1+ cos 56 = 2 cos^2 28 as cos 2t = 2 cos^2 t - 1
cos 58 = 2 cos^ 29 -1
cos 66 = 2cos ^2 33 -1

so 1+ cos 56 + cos 58 - cos 66
= 2 cos^2 28 + 2 cos^2 29 - 2 cos^2 33
= 2 ( cos^2 28 + cos ^2 29 - cos ^2 33)

now 28+29+33 = 90

if we prove cos ^2 A + cos^2 B - cos^2 c = 2 cos A cos B sin C when A+B+C = pi.2
then we are through

2008/008) Tough inequality

Prove that for all x>0, y>0, and all real a it holds true that

(x^((sin(a))^2))*(y^((cos(a))^2))
let us assume y > x

x^t y^(1-t) < x+ y putting (sin a)^2 = t we get (cos a)^2 = 1-t

devide by x on both sides

x^(t-1)y^(1-t) < (1+y/x)

or (y/x)^(1-t) < 1+ y/x

or m ^k < 1+m where k < 1 and m > 1
we know m ^k < m where k < 1

so m^ k < 1+m

if x > y then role of sin a and cos a are reversed and you get the same result

2008/007) Verify (tanx+secx)/(secx-cosx+tanx) = cscx

AS tan x + sec x is in both numerator and dnominator and denominator has one extra term I would take reciprocal of LHS

1/ LHS = 1 - cosx/(tan x + sec x)
= 1- cos x ( sec x- tan x)/ (tan x + sec x)( sec x- tan x)
= 1 - cos x ( sec x - tan x)/ (sec^2 x - tan ^2 x)
= 1- cos x( sec x - tan x) as sec^2x - tan ^2x = 1
= 1 - cos x(1/cosx - sinx /cos x)
= 1-1+ sin x = sin x
so LHS = 1/ sin x = csc x


proved

2008/006) If z = cos x + i sin x ,then simplify (z^2 -1)/(z^2 +1)

(z^2-1)/(z^2+ 1) = (z - 1/z)/ (z+ 1/z) ...1

now z = cos x + i sin x ... 2

so z = e^ix

so 1/z = e ^-ix = cos x - i sin x ... 3

add 2 and 3 to get z+ 1/z = 2 cos x

subtract 3 from 2 to get

(z-1/z) = 2i sin x

from 1 by deviding (z^2-1)/z^2+1)= 2i sin x/(2 cos x) = i tan x

2008/005) rational no. p,q and r satisfying the property that pq+qr+rp=1

rational no. p,q and r satisfying the property that pq+qr+rp=1
prove that (p^2+1)(q^2+1)(r^2+1) square of a rational number

so p = (1-qr)/(q+r)
if we chose q = tan A and r = tan B

we get
1/p = (q+r)/(1-qr) = (tan A + tan B)/(1- tan A tan B) = tan (A+B)

or p = cot (A+B)

now
(p^2+1) (q^2+1)(r^2+ 1) = sec^2 A sec ^2 B cosec^2 (A+B)
= (sec^2 A sec^B)/ sin^2 (A+B)

this is square of reciprocal of sin (A+B) cos A cos B
sin (A+B) cos A cos B
=( sin A cos B + cos A sin B)cos A cos B
= sin A cos A cos ^2B + cos^2 A sin B cos B
= tan A cos ^2 A cos ^2 B + tan B cos ^2 A cos ^2 B
= (tan A + tan B)/(sec^2 A sec ^2B)
= (tan A + tan B)/(1+ tan ^2 A)(1+ tan ^2B)

so (p^2+1) (q^2+1)(r^2+ 1) = ((1+ tan ^2 A )(1+tan ^2B)/(tan A + tan B))^2

if tan A and tan B that is q and r are rational then

((1+ tan ^2 A )(1+tan ^2B)/(tan A + tan B)) is rational and so (psquare +1)(qsquare +1)(rsquare +1) is the square of a rational number

2008/004) If α + β + γ = π/2, show that,

If α + β + γ = π/2, show that,

[(1 - tan α/2)(1 - tan β/2)(1 - tan γ/2)]/[(1 + tan α/2)(1 + tan α/2)(1 + tan α/2)]
= (sin α + sin β + sin γ - 1)/(cos α + cos β + cos γ)

proof:
We should start with the RHS as it is more complex

Before we proceed let us use/deduce certain information as they shall be used
(actually they should be derived as required but I am deriving to keep the flow
α + β + γ = π/2
so α + β = π/2 – γ
or α + β = π/2 – γ …1

cos a + cos b = 2 cos (a+b)/2 cos(a-b)/2 … 2

taking sin of both sides of 1 we get
sin (α + β) = sin (π/2 – γ) = cos γ … 3
sin 2a = 2 sin a cos a …. 4

from 1
(α + β)/2 = (π/2 – γ)/2
So sin (α + β)/2 = sin (π/2 – γ)/2 = cos (π/2 + γ)/2 … 5

Again as
α + β + γ = π/2
so α - β + γ = π/2 - 2 β
so α - β + γ + π/2 = π - 2 β
so (α - β + γ + π/2)/ 4 = π/4 - β/2 …6

Again as
α + β + γ = π/2
so α - β - γ = π/2 - 2 β – 2 γ
so α - β - γ - π/2 = - 2 β – 2 γ
so (α - β - γ - π/2)/ 4 = - (β + γ)/2
so cos (α - β - γ - π/2)/ 4 = cos (β + γ)/2 as cos - A= cos A
= cos (π/2 – α)/2
= cos (π/4 – α/2) .. 7
And sin (α - β - γ - π/2)/ 4 = - sin (β + γ)/2 = - sin (π/4 – α/2)
Or sin (-α + β +γ + π/2)/ 4 = sin (π/4 – α/2) … 8


Now let us find he denominator
cos α + cos β + cos γ
= 2 cos (α + β)/2 cos(α - β )/2 + cos γ (using 2)
= 2 cos (α + β)/2 cos(α - β )/2 + sin (α + β) (using 3)
= 2 cos (α + β)/2 cos(α - β )/2 + 2 cos (α + β)/2 sin (α + β)/2 ( using 4)
= 2 cos (α + β)/2 (cos(α - β )/2 + sin (α + β)/2)
= 2 cos (π/2 - γ)/2 (cos(α - β )/2 + cos (π/2 + γ)/2)) (using 1 and 5)
= 2 cos (π/4 – γ/2)*2 cos (α - β + π/2 + γ)/2 cos (α - β - π/2 - γ)/2 (using 2)
= 4 cos (π/4 – γ/2 ) cos (π/4 - β/2) cos (α - β - π/2 - γ)/2 (using 6)
= 4 cos (π/4 – γ/2 ) cos (π/4 - β/2) cos (π/4 – α/2) using 7

So cos α + cos β + cos γ = 4 cos (π/4 – γ/2 ) cos (π/4 - β/2) cos (π/4 – α/2) … (A)
Now numerator

For the additional identities

Sin A + sin B = 2 sin (A+B)/2 cos (A-B)/2 … 9
sin (α + β + γ) = sin π/2 = 1 …. 10

sin A – sin B = 2 sin (A-B)/2 cos (A+B)/2 … 11


cos A – cos B =2 sin (A+B)/2 sin (B-A)/2 … 12

tan (π/4- A)= (1- tan π/4 tan A) / (tan π/4 + tan A) = ( 1- Tan A)/(1+tan A) … 13

based on above numerator

sin α + sin β + sin γ – 1
= 2 sin (α + β)/2 cos(α - β )/2 + sin γ – sin (α + β + γ) (using 9 and 10)
= 2 sin (π/4 – γ/2) cos(α - β )/2 –(sin (α + β + γ) - sin γ) (using 5)
= 2 sin (π/4 – γ/2) cos(α - β )/2 - 2 sin (α + β)/2 cos (α + β + γ) + γ)/2 (using 11)
= 2 sin (π/4 – γ/2) cos(α - β )/2 - 2 sin (π/4 – γ/2) cos (π/2+ γ)/2 (using 5)
= 2 sin (π/4 – γ/2) (cos(α - β )/2 - cos (π/2+ γ)/2)
= 2 sin (π/4 – γ/2) (2 sin (α - β + π/2+ γ)/4 sin (-α + β +π/2-+γ)/4 (using 12)
= 4 sin (π/4 – γ/2) sin (π/4 – β/2) sin (π/4- α/2) (using 6 and 8)

sin α + sin β + sin γ – 1 = 4 sin (π/4 – γ/2) sin (π/4 – β/2) sin (π/4- α/2) …(B)

from A and B we get
RHS =
(sin α + sin β + sin γ – 1)/ (cos α + cos β + cos γ)
= (4 sin (π/4 – γ/2) sin (π/4 – β/2) sin (π/4- α/2))/ (4 cos (π/4 – γ/2 ) cos (π/4 - β/2) cos (π/4 – α/2))
= tan (π/4 – γ/2) tan (π/4 – β/2) tan (π/2- α/2)
= tan (π/4- α/2) tan (π/4 – β/2) tan (π/4 – γ/2)
= [(1 - tan α/2)(1 - tan β/2)(1 - tan γ/2)]/[(1 + tan α/2)(1 + tan β/2)(1 + tan γ/2)]/[( (using 13 and rearranging the terms)

= LHS