Question

50 points !!


solve only if u know
wrong answers will be reported !!!​

Answer 1

We know that,

cos2x= (1-tan^2x)/(1+tan^2x)

cosx= (1-tan^2(x/2))/(1+tan^2(x/2))

sin2x= (2tanx)/(1+tan^2x)

sinx=(2tan(x/2)/(1+tan^2(x/2))

To prove:

tan^-1(cosx/1+sinx)= π/4-x/2, x€(-π/2,π/2)

Proof:

LHS:

tan^-1(cosx/1+sinx)

tan^-1((1-tan^2(x/2)/1+tan^2(x/2))/ (2tan(x/2)/1+tan^2(x/2)))

tan^-1(1-tan^2(x/2)/1+tan^2(x/2)+2tan(x/2))

Applying formula a^2-b^2=(a+b)(a-b)

tan^-1(1+tan(x/2)(1-tan(x/2)/(1+tan(x/2)^2)

tan^-1(1-tan(x/2)/1+tan(x/2)) --------------(1)

tan(x-y)= tanx-tany/1+tanxtany

tan(45-x)=tan45°-tanx/1+tan45°tanx

tan45°=1

tan(45°-x)=1-tanx/1+1×tanx

Applying same concept in eqn (1) we get

tan^-1(tan(45°-(x/2))

tan^-1 and tan will cancel out ,

45°-x/2

45° in radian is π/4.

(Formula for converting 45° to radian π×45°/180=π/4)

Therefore LHS

π/4-x/2 where x€(-π/2,π/2)

RHS

π/4-x/2 where x€(-π/2,π/2)

Hence Proved

Answer 2

tan-1[cosx / (1+sinx)]

=tan-1[ sin(π/2 - x) / (1+cos(π/2 - x))]

Let t=π/2-x

=tan-1[ sint / (1+cost)]

=tan-1[2sin(t/2) cos(t/2) / (2 cos2(t/2))] {since sin2x=2sinx cosx and 1+cos2x=2cos2x }

=tan-1[sin(t/2)/cos(t/2)]

=tan-1[tan(t/2)]

=t/2

=(π/2 - x)/2

=π/4 - x/2

Q.E.D

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