Question

Good after noon

Please solve 15th , 18th , and 22nd .........

Answer 1

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15.


Take the LHS 
We know tan 2x = 2tanx/( 1-tan²x) 

Using this we can write, 
tan 4A = tan2(2A) 

= 2 tan2A/( 1- tan² 2A)............(1) 

tan2A = 2tanA / ( 1- tan² A)...............(2) 

Putting (2) in (1) 

2 [2tanA / (1- tan² A)] 
= ------------------------------------
1- [ 2 tanA / (1- tan²A)]² 

4 tanA /( 1- tan²A) 
= ------------------------------------------------
[ (1- tan²A)² - 4tan²A] /( 1-tan²A)² 

[Cross multiply in denominator , now one of 1- tan²A gets cancelled] 

4tan A 
= --------------------------------------------------------------
[1 + tan^4 x - 2tan² A- 4tan²A]/(1-tan²A) 

Denominator of the denominator is the numerator 

= 4tanA ( 1- tan²A)/[1 - 6tan²A + tan^4A] 

= (4 tanA - 4 tan³A) /[ 1- 6tan²A + tan^4A] 








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18.




Prove that : tan(45+A)+tan(45-A)=2sec2A? 

L.H.S. = tan(45+A)+tan(45-A) = 

= [tan(45)+tanA] / [1-tan45.tanA] + [tan(45)-tanA] / [1+tan45.tanA] = 

= (1+tanA) / (1-tanA) + (1-tanA) / (1 + tanA) = 

= (1+tanA)^2 / (1-tan^2A) + (1-tanA)^2 / (1 - tan^2A) = 

= [(1+tanA)^2 + (1-tanA)^2] / (1-tan^2A) = 2(1+tan^2A) / (1-tan^2A) = 

= 2sec2A = R.H.S. >===============================< ANSWER





22.

LHS = Cos2x + Cos2(x+π/3) + Cos2(π/3 -x)

  = (1+cosx)1/2 + (1+cos(x+^/3))1/2 + (1+cos(π/3-x)) 1/2 

  = 1/2 ( 1+cosx + 1 +Cos(x+^/3) + 1 + Cos(π/3-x))

  = 1/2 ( 3 + cosx + Cos( x + 60 ) + Cos( 60 - x )

.  = 1/2 ( 3 + Cosx + 2 cos120. cosx)

  = 1/2 ( 3 + cosx + 2 cosx . -1/2) 


(since  cos 120 = -1/2)

 
= 1/2 ( 3 + cosx - cosx)

 
= 3/2 = RHS

Answer 2

hope it help you my bestie !