Prove: sec^2x + csc^2x = sec^2 * csc^2x

proof:
sec^2x + csc^2x
= sec^2x ( 1+ csc^2 x/ sec^2 x)
= sec^2x ( 1+ 1/sin ^2 x /(1/cos ^2x ))
= sec^2 x( 1+ cot^2 x)
= sec ^2 x csc^2 x

Prove (1 - cosX) / (1 + coxX) = (cscX - cotX)^2

to remove 1+ cos x from denominator multiply numerator and denominator by 1- cos x

= (1-cos x)^2/((1+ cos x)(1-cos x))
= (1- cos x)^2/ (1- cos^2 x)
= (1- cos x)^2/ sin ^2 x
= ((- cos x)/sin x)^2
= (csc x - cot x)^2

How to integrate ∫(cos3x)^2?

using cos 2y = 2 cos^2y - 1

you get putting y = 3x

cos 6x = 2 cos ^2 3x - 1

or cos^2 3x= (cos 6x + 1)/2

now knowing integral of cos nx to be sin nx/n you get integrating

(1/6 sin 6x)/2 + 1/2 x

the constant of integration need to be added to get 1/12 sin 6x + x/2 + C

Prove :64 {Cos^8(x) + Sin^8(x)} = cos8 x + 28cos 4x + 35

we know (cos^2 x + sin ^2 x) = 1

square both sides


cos^4 x + sin ^4 x + 2 cos^2 x sin ^2 x = 1

or cos^4 x + sin ^4 x = 1- 2 cos^2 x sin ^2 x

multiply by 2 to get

2(cos^4 x + sin ^4 x) = 2- 4 cos^2 x sin ^2 x = 2 – ( 2 sin x cos x)^2 = 2 – sin ^2 2x

multiply by 2 again to get

4(cos^4 x + sin ^4 x) = 4 – 2 sin ^2 2x = 3 + cos 4x

again square both sides to get

16(cos^8x + sin ^ 8x+ 2 cos ^4 x sin ^4 x) = 9 + cos^2 4x + 6 cos 4x

or 16(cos ^ 8x + sin ^8x ) = 9 + cos^2 4x + 6 cos 4x – 32 cos ^4 x sin ^4 x
= 9 + cos^2 4x + 6 cos 4x – 2* (2 cos x sin x)^4
= 9 + cos^2 4x + 6 cos 4x – 2* (sin 2x)^4
multiply by 4 on both sides

64(cos ^ 8x + sin ^8x) = 36 + 4 cos^2 4x + 24 cos 4x – 2* (2sin^2 2x)^2
= 36 + 4 cos^2 4x + 24 cos 4x – 2* (1- cos 4x)^2
= 36 + 4 cos^2 4x + 24 cos 4x -2(1- 2cos 4x + cos^2 4x)
= 34 + 2 cos^2 4x + 28 cos 4x
= 34 + (1+ cos 8x) + 28 cos 4x
= 35 + cos 8x + 28 cos 4x

prove that sin10*sin50*sin60*sin70*sin90 =root3/16

proof
we know sin 90 = 1 and sin 60 = sqrt(3)/2

hence
sin10*sin50*sin60*sin70*sin90 = sqrt(3/2) sin 10 sin 50 sin 70
= sqrt(3)/2 sin 10 sin (90-40) sin (90-20)
= sqrt(3)/2 sin 10 cos 40 cos 20 ( as sin (90-x) = cos x)
= sqrt(3)/2 sin 10 cos 20 cos 40 ..1

Knowing sin 2a =2 sin a cos a and 20 is double of 10 and 40 is double od 20 we proceed

now sin10 cos 20 cos 40 = (cos 10 sin 10 cos 20 cos 40)/cos 10
= (2 cos 10 sin 10 cos 20 cos 40)/(2 cos 10)
= ( sin 20 cos 20 cos 40)/( 2 cos 10
= (2 sin 20 cos 20 cos 40)/( 4 cos 10)
= ( sin 40 cos 40)/( 4 cos 10)
= ( 2 sin 40 cos 40)/( 8 cos 10)
= sin 80 /( 8 cos 10)
= cos 10/(8 cos 10) = 1/8 ...2

from 1 and 2 we get the result

prove that 1 - sin x / 1 + sin x = tan^2(π/4 - x/2) ,if 0≤X≤π/2

we have (sin A + cos A)^2 = sin ^2 A + 2 sin A cos A + cos ^2 A = 1 + 2 sin A cos A

(sin A - cos A)^2 = sin ^2 A - 2 sin A cos A + cos ^2 A = 1 - 2 sin A cos A

tan (pi/4-A) = (tan pi/4 - tan A)/(1 + tan pi/4 tan A) = (1- tan A )/(1+ tan A)

LHS =
(1 - sin x) / (1 + sin x)
= (1- 2 sin x/2 cos x/2)/(1+ 2 sin x/2 cos x/2)
= ( sin x/2- cos x/2)^2/(sin x/2+ cos x/2)^2 (from above)
= ((sin x/2- cos x/2)/ (sin x/2 + cos x/2))^2
= ((1- tan x/2)/(1+ tan x/2))^2
= ( tan (pi/4-x/2))^2 (from above)

proved

Find all x, 0 < x < pi, such that sin^2x=1/2

sin x = 1/ sqrt(2) or - 1/sqrt(2)

x= pi/4, 3pi/4 , 5pi/4 or 7pi/4 (1st 2 for 1/sqrt(2) and next 2 for - 1/sqrt(2)

If A, B, C are the angles of a triangle...?

prove that cos [(B-C) / 2] - sin (A/2) = 2 sin (B/2) sin (C/2)

A+B+C = pi

so A/2 = pi-(B+C)/2

sin (A/2) = sin (pi-(B+C)/2) = cos (B+C)/2

so
cos [(B-C) / 2] - sin (A/2) =

= cos [(B-C) / 2] - cos [(B+C)/2] = 2 sin (B/2) sin (C/2)

using COS A - COS B = 2 cos sin (A+B)/2 sin (A-B)/2

simplify the following?

simplify the following? cos(2x-y)cosy - sin(2x-y)siny?

using cos A cos B - sin A sin B = cos(A+B) and putting A = 2x-y and B = y we get

cos(2x-y)cosy - sin(2x-y)siny = cos (2x-y+y) = cos 2x

prove that (1-sinx)/cosx+cosx/(1-sinx)=2sec x

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

prove that (tanx-cotx)/(tanx+cotx)=1-2cos^2x

LHS =
(tanx-cotx)/(tanx+cotx))
= (tan x- 1/tan x)/(tan x+ 1/tan x)
= (tan ^2 x-1)/(tan^2 x+ 1)
= (tan ^2 x-1)/ sec^2 x
= cos^2 x( tan ^2 x - 1)
= sin ^2 x - cos ^2 x
= (1- cos^2 x) - cos^2 x = 1- 2 cos^2 x
= RHS proved

verify the identity (tan x – sec x + 1)/(tan x + sec x - 1) = cos x(1 + sin x)?

as sec^2 x = tan ^2 x + 1 so 1 = sec^2 x - tan ^2 x

so putting 1 = sec^2 x- tan ^2 x in denominator we get

LHS -= (tan x – sec x + 1)/(tan x + sec x - 1)
= (tan x – sec x + sec^2 x- tan ^2 x)/(tan x + sec x - 1) replacing 1 with sec^2 x - tan ^2 x in numerator
= (tan x – sec x + (sec x- tan x)( sec x + tan x))/(tan x + sec x - 1) factoring sec^2 x - tan ^2 x
= (sec x- tan x)(-1 + sec x + tan x)/(tan x + sec x -1) taking sec x - tan x common
= sec x - tan x canceling common part
= 1/cos x - sin x/ cos x
= (1- sin x)/cos x
= cos x(1- sin x)/cos^2 x( multiplying numerator and denominator by cos x)
= cos x(1- sin x)/(1- sin ^2 x)
= cos x(1- sin x)/(1- sin x )(1+ sin x)
= cos x/(1+ sin x)
proved

Prove cos x /(1+sin x) + ((1+sin x )/cos x ) = 2 sec x

cos x/(1+sin x) + (1+sin x/cos x
= cos x(1- sin x)/((1+ sin x)(1- sin x)) + (1+sin x)/cosx
= cos x (1- sin x)/(1- sin ^x) + (1+sin x)/cos x
= cos x(1- sin x)/cos^2 x) + (1+sin x)/cos x
= (1- sin x)/cos x + (1+sin x)/cos x
= (1- sin x + 1 + sin x)/ cos x
= 2/ cos x
= 2 sec x